Q reference count

Is there a way to tell how many references to a specific named Q have been attributed? For example, if an application uses the Queue.vi get three times to create the queue that is referenced by, I would like to be able to query the number of active queues (for this unique name) that have been created.

Thank you

-cb

I am sure that you can not get this information.

Instead, you can limit the creation and waste out of the queue to a manager of VI, which will also keep track of the arbitrators, but nothing prevents someone else, unless you wrap all interactions with the queue in a class.

Tags: NI Software

Similar Questions

I have a PC brand new - HP and played Chess Titans. Now all of a sudden it says it cannot find the texture, but he said no zero the reference count and then could not find device 3D.

original title: Chess Titans

I have a PC brand new - HP and played Chess Titans. Now all of a sudden it says it cannot find the texture, but he said no zero the reference count and then could not find device 3D. What's past and how to fix this. Thank you.

These error messages can occur if you (or software) also disabled Aero. Aero must be on (and supported by the video card) to play Chess Titans. "192GO should be enough for everyone." (of the miniseries "Next generation jokes")

Differences in size and HD file size

I have two questions.

I. how a 1 TB with 925,67 GB disk volume...

... contains a record of Time Machine Backups.backupdb 4.55 TB?

II. with more than a dozen other disk images (.dmg) in the same folder of your choice, why there disk utility chose this one to treat separately?

Time Machine uses the file system "physical" which, if the software does not know how to treat hard links will double, triple, quadruple, etc. have the same file.

A hard link is a 2nd, 3rd, 4th, etc... name of a directory that points to the same exact physical file. Whenever a hard link is created, the file gets a reference count incremented. A directory entry for the file is removed every time, the hardlink reference count is decremented. Then the account goes to zero, the data file are deleted.

Back to TimeMachine. TimeMachine makes it look like you have several complete your file system copies, but in reality if a file is unchanged in 2 adjacent backups, a permanent link will be put into the current backup pointing to the previous identical file. Now it seems like backup have a copy of the file, but there is ONLY 1 copy of the data on the disk. If the file has been changed, then the current backup gets its own copy of the new or modified file and no hard link is created for this file during the backup.

It is to try to understand a storage count app that walks the directory had a file with hard links 2 times or more significant. The app must keep track of each file identifier he saw and then check each file that is looking against the previous files, he saw to see if they have the same file identifier. It's not bad for a small number of files, but when you start talking about tons of files (in millions), you start to need a lot of memory to keep a record of each found file identifier, and you need fast search algorithms which allows to quickly add new items other than the software metering will eventually slow down.

However, to the best of my knowledge the Finder does not track of hard links. Especially because very few users actually create them, and few people to get information from a Time Machine backup.

Now the volume itself maintains counters of allocated and free space, so it is not confused by the harlinks, and given that these values are updated each time storage is allocated or freed, it's quick return this information at the request of Volume Get Info.

II. with more than a dozen other disk images (.dmg) in the same folder of your choice, why there disk utility chose this one to treat separately?

I do not understand the question. Then again, maybe I would not have understood the question if you have changed the wording. I do not look that closely at utility disc info.

Get the 1097 error when trying to call a Labview DLL

I had to write a wrapper DLL to use some functions in the DLL to a third party, and I'm having a few problems. I created an executable to test my code and it works fine, but when I try to execute the same function in the DLL that I created in Labview, I get error 1097. I read through most of the posts on this forum about this issue, but none of the proposed solutions are working for me. I'm sure this isn't a question of transfer settings to COLD LAKE because I changed the function so it only accepts a single entry (a number he obtains from a Subvi, which came from the third-party DLL, this entry is working with a bunch of other functions, so I don't think that is the question).

I have attached the C code that creates the wrapper, but I don't know how it is useful without familiarity with the functions of the original DLL. I hope that someone can spot something basic I am out of my code, but please let me know if I need to call something else.

I use a Labview 32 bit version 2012 on a 64-bit Windows 7 computer.

Thanks in advance for any help!

You should read the documentation on LoadLibrary() and understand. Essentially if you provide a full path to LoadLibrary() the DLL MUST be present at this place so that LoadLibrary() will succeed. If you pass only the name Windows DLL will ONLY search in these places:

(1) if the module (the DLL) is already loaded it will return a handle to the loaded module and increment its reference count

(2) if the referencing module contains a manifest (either incorporated into the image module or in a separate manifest file in the same directory as the executablemodule) specyfying a version number for this DLL will load this DLL with this version of the SxS (side by side) location and failure if it is not present

(3) if the executable load contains a manifest (either incorporated in the executable file or in a separate manifest file in the same directory as the executable file) number specyfying a version for this DLL, it will load this DLL with this version of the location of SxS (side by side) and fail if it is not present

(4) it will search for the DLL in the same directory as the directory where the current image of the process is loaded. It is NOT the directory of the DLL loading the DLL, BUT the directory where the EXE file from which the process was created.

(5) in the system directory

(6) in the Windows directory

(7) any directory in the PATH environment variable

(8) the path of current directory which is a location managed by windows by process and changed the two explicitedly when the application calls the API SetCurrentDirectory() but also implicitedly by things like the file selection dialog whenever it is rejected by other means than the undo operation.

Essentially putting your dependend DLL in the same directory as the wrapper DLL does NOT work when you use the name of the DLL only UNLESS you place the two DLLs in the same directory as your executable file. If you don't want this restriction and want to be able to move these modules together in different directories, you don't have to create the correct path in your wrapper to LoadLibrary(). Basically, you must have the code to determine the location of your DLL wrapper path and then strip her own name and add the DLL name wrapped in it and then move to the LoadLibrary().

There are several ways to determine the path of the current module. The simplest if you don't mind having the name hard-coded in your code DLL would be to ask GetModuleFileName (GetModuleHandle ("yourDLLname"), buffer, size).

buffer allocation and minimizing memory allocation

Hello

I am tryint to minimize the buffer allocation and memory in general activity. The code will run 'headless' on a cRIO and our experience and that of the industry as a whole is to ellliminate or minimize any action of distribution and the dynamic memory deallocation.

In our case we treat unfortunately many string manipulations, thus eliminating all the alloc/dealloc memmory is significant (impossible?).

Which leaves me with the strategy of "minimize".

I did some investigation and VI of profiling and play with the structure "on the spot" to see if I can help things.

For example, I have a few places where I me transpoe a few 2D charts. . If I use the tool 'See the buffer allocations' attaced screenshot would indicate that I am not not to use the structure of the preliminary examination International, both for the operation of transposition of the table for the item index operations? As seems counter intuitive to me, I have a few basic missunderstanding either with the "show stamp" tool of the preliminary examination International, or both... The tool shows what a buffer is allocated in the IPE and will once again out of the International preliminary examination, and the 2D table converts has an allowance in and out, even within the IPE causing twice as many allowances as do not use REI.

As for indexing, using REI seems to result in 1.5 times more allowances (not to mention the fact that I have to wire the index numbers individually vs let LabVIEW auto-index of 0 on the no - IPE version).

The example illustrates string conversions (not good from the point of view mem alloc/dealloc because LabVIEW does not determine easily the length of the 'picture' of the chain), but I have other articles of the code who do a lot of the same type of stuff, but keeping digital throughout.

I would be grateful if someone could help me understand why REI seems to increase rather than decrease memory activity.

(PS > the 2D array is used in the 'incoming' orientation by the rest of the code, so build in data table to avoid the conversion does not seem useful either.)

QFang wrote:

-My reasoning (even if it was wrong) was to indicate to the compiler that "I do not have an extra copy of these tables, I'll just subscribe to certain values..." Because a fork in a thread is a fairly simple way to increase the chances of duplications of data, I thought that the function index REI, by nature to eliminate the need to split or fork, the wire of the array (there an in and an exit), I would avoid duplication of work or have a better chance to avoid duplication of work.

It is important to realize that buffer allocations do occur at the level of the nodes, not on the wires. Although it may seem to turn a thread makes a copy of the data, this is not the case. As the fork will result in incrementing a reference count. LabVIEW is copy-on-write - no copy made memory until the data is changed in fact, and even in this case, the copy is performed only if we need to keep the original. If you fork a table to several functions of Board index, there is always only one copy of the table. In addition, the LabVIEW compiler tries to plan operations to avoid copies, so if several branches read from a wire, but only it changes, the compiler tries to schedule the change operation to run after all the readings are made.

QFang wrote:

After looking at several more cases (as I write this post), I can't find any operation using a table that I do in my code that reduces blackheads by including a preliminary International examination... As such, I must STILL understand IPE properly, because my conclusion at the present time, is that haver you 'never' in them for use. Replace a subset of a table? no need to use them (in my code). The indexing of the elements? No problem. .

A preliminary International examination is useful to replace a subset of the table when you're operating on a subset of the original array. You remove the items that you want, make some calculations and then put back them in the same place in the table. If the new table subset comes from somewhere other than the original array, then the POI does not help. If the sides of entry and exit of International preliminary examination log between them, so there no advantage in PEI.

I am attaching a picture of code I wrote recently that uses the IPEs with buffer allocations indicated. You can see that there is only one game of allowances of buffer after the Split 1 table D. I could have worked around this but the way I wrote it seemed easier and the berries are small and is not time-critical code so there is no need of any optimization. These tables is always the same size, it should be able to reuse the same allowance with each iteration of the VI, rather than allocate new arrays.

Another important point: pads can be reused. You might see a dot of distribution on a shift register, but that the shift register must be assigned only once, during the first call to the VI. Every following call to the VI reuses this very spot. Sometimes you do not see an allocation of buffer even if it happens effectively. Resizing a table might require copying the whole table to a new larger location, and even if LabVIEW must allocate more memory for it, you won't always a point of buffer allocation. I think it's because it is technically reassign an existing table instead of allocating a new, but it's always puzzled me a bit. On the subject of the paintings, there are also moments where you see a point to buffer allocation, but all that is allocated is a 'subfield' - a pointer to a specific part of an existing table, not a new copy of the data. For example 1 reverse D table can create a sub-table that points towards the end of the original with a 'stride' array-1, which means that it allows to browse the transom. Same thing with the subset of the table. You can see these subtables turning on context-sensitive help and by placing the cursor on a wire wearing one, as shown in this image.

Unfortunately, it isn't that you can do on the string allocations. Fortunately, I never saw that as a problem, and I've had systems to operate continuously for months who used ropes on limited hardware (Compact FieldPoint controllers) for two recordings on the disk and TCP communication. I recommend you move the string outside critical areas and separate loop operations. For example, I put my TCP communication in a separate loop that also analyses the incoming strings in specific data, which are then sent by the queue (or RT-FIFO) to urgent loops so that these loops only address data of fixed size. Same idea with logging - make all string conversions and way of handling in a separate loop.

Sampling rate higher for the measurement of precision meter

I have a BNC 6259 M Series DAQ USB. I am currently using the DAQ Assistant to perform simple cash rising measured with the measuring mode single sample on request. I tested my VI with a known square wave with a function generator signal and it clearly lacks a few edges. I think that the solution lies in faster sampling. However, I was not able to understand how to use clocks to set up continuous sampling mode. I tried the forums and I found articles that were close, but not quite exactly the problem I am facing with as (http://forums.ni.com/t5/Digital-I-O/trying-to-use-NI-6251-s-DIO-port-as-input-and-output/m-p/448035#...) or (http://forums.ni.com/t5/Multifunction-DAQ/Using-Counter-of-PCI-6024E-with-Quadrature-Encoder/m-p/984...). Any guidance here would be great.

In fact, the calendar should not have anything to do with the edges being detected. Configuration of a sample clock for a county of just edge task allows you to enjoy deterministically in the account register and has no impact on the edges which can be counted. Also, the analog examples really have nothing to do with what you seem to be asking questions on.

... So it leaves the question unanswered as to why you might miss the edges. Perhaps the following information could shed some light on the question:

1. it is possible that you do not configure the counter exactly as you think you are. Can you post the VI you use? As a point of reference, count digital events shipping example does not use any clock sample timing and just questioned the value of the register count with a software loop, but the meter should not miss all this edge on the input source. There are examples that are timed by the material available as well, but this is not necessary, unless you need a constant specified dt between your counter samples. To use the examples of the timed sample, you will need to generate a clock of either another subsystem on the map or use an external clock.

2. assuming that the configuration of the counter is not the issue, there may be a problem with the method that you use to determine if you are away from the edges. How do you know that you are away from the edges? The function generator produces only a finite pulse amount? You start the meter before start out impulses?

3. If the two points above do not raise red flags, it seems likely that the meter is registered just not some of the impulses of your FGEN. Can I assume that the output of the FGEN is 0 - 5V TTL? What is its frequency and duty cycle? The maximum external source for the meter on the M-series products: DAQ (like the 6259) is specced at 20 MHz, but this depends on a clean signal with good connections. At frequencies above it, the bandwidth of the front-end of the PFI lines becomes limiting. If you have an available specification document for your FGEN I'd like to be able to see it.

I hope this gets you throw on the right track to solve the problem - impatience comes back with more information.

Best regards

Release of an object created with ActiveXAdapter

Hello

I have an ActiveX (HyperAccess) object that uses the ActiveXAdapter.

Is that what I created during execution. Communication works very well.

But I'm not able to call ReleaseDispatch. His is not accessible the Hilgraeve interface.

It is a way to release the ActiveX (IDispatch) objects during execution?

Jürgen

Fair value of all variables storing the reference to nothing.

Example:

Locals.MyActiveXReference = Nothing

Reference count is automatically handled by the object reference variables. If you step to the activeX to create the object at the time of load/unload you must unload the not too well (but is not the default setting).

Hope this helps,

-Doug

Question about Equallogic snapshots.

Hello forum

I could not confirm this, but instant never need to commit anything to the basic volume?

Lets say I have 10 snapshots and delete the oldest? There will be a load of e/s to validate these instant changes?

Also, if someone could point me to any type of indepth explanation of this technique of snapshot, I would be interested to read all that.

Thank you!

N ° current data are already on the base volume. There is nothing to commit. Only if you are completely restoring a volume from a snapshot will be there in writing for the base volume.

On the site of Equallogic Support KB:

Solution title TABLE: how the snapshot reserve space is allocated and used

Solution details Snapshot Reserve Allocation and use

-------------------------------------

In a series group PS, snapshots can protect against errors, of a volume

virus or the database. A cliché represents the content of a volume

at the time you created the snapshot. Creating a snapshot does not disturb

access to the volume, and the snapshot is instantly available allowed

iSCSI initiators.

Before you can create snapshots for a volume, you must allocate space (called

snapshot pool) for snapshots. You set the value of the snapshot reserve

When you create or modify a volume. Snapshot reserve benefits from the same

as the volume of the pool data.

When snapshot data consume entire supply snapshot, the Group remove is

the oldest snapshots to free up space for new images or sets the volume and

snapshots offline, according to the policy you selected for instant recovery

space.

The functionality for creating snapshots is called hybrid allocate when writing.

Operation of sharing and snapshot of the page

-----------------------------------

A PS Series Group organizes the physical storage into logical segments called pages.

Each page is composed of several logical blocks. This is similar to the way file

systems combine areas of physical disk in 'clusters' or 'chunks '. Each page has

a reference count which records the number of volumes and snapshots that share

the page.

When you create a volume, the group creates an internal table, called the volume

table, which contains pointers to pages that use the volume. When you create

a snapshot of a volume, the group creates a snapshot table by making a copy of

the volume table, which is usually an operation in memory that the Group

run in a transactional manner. For each page of volume in use, the support group increases the

reference count to indicate that the volume and the snapshot share page.

Because the group does not move or copy the data or allocates a new pages, photos

are fast and efficient.

Reserve of the snapshot stores the differences between the data on the volume and snapshot data

(in addition to differences between the data of multiple snapshots). When you

first create a snapshot, the page for the volume and the snapshot tables are

In brief identical copies and the snapshot consumes no snapshot

reserve. A reading of the same logical block of the volume application or the

snapshot returns the same data because the application is reading from the same page.

However, if you write a page that has a volume and a snapshot of share, snapshot

reserve is consumed.

Here's a simplified example of a snapshot operation. In general, no.

additional I/o operations are needed to manage the data volume or snapshot.

However, other internal operations can occur due to virtualization and

data balancing on berries of the PS Series.

If an application performs a write to 8 KB for a volume containing a snapshot, the

Group:

1. determine what page is modified by the write operation.

2 - If the page is not shared, writes the data to the page.

3. If the page is shared:

. (a) allocates a new page of disk space and reduces the instant to reserve

. the volume of a single page.

. (b) update the page of volume table to point to the newly allocated page.

. (c) mark the newly allocated page as having new data on the volume and the references of the

. original page for unchanged data.

. (d) writes the data to the new page.

When writing is complete, if you read the data on the volume, you have access to the

new page and new data. However, if you read the same logical block of the

Instant, you get the original data, because the snapshot will always point towards the

original page. Similarly, if you set a snapshot online, write to the snapshot.

feature hybrid write protects the original data volume by allocating

a new page for the new snapshot data.

Only the first page of writing to a volume shared (or snapshot) consumes additional

snapshot reserve. Each subsequent entry is considered identical to a writing on a

non-shared the page because the original data are already protected.

Functionality similar to hybrid allocate when writing is used in cloning operations.

However, unlike when you create a snapshot, cloning a volume immediately

consumes space additional group. If a clone is moved to another pool, data

is copied during the operation of moving pool.

Restoring a Volume from a snapshot

----------------------------------

Because of the layout table, restore a volume from a snapshot is very

quick. First of all, the group automatically creates a snapshot of the volume by copying

the volume table to a new table of snapshot. Then the Group transposes the page tables

the volume and the snapshot you selected for the restore operation. NO.

additional space is required, and no data is moved.

Deletion of Volumes and Snapshots

------------------------------

Because volumes and snapshots to share pages, if you delete a volume, you

automatically remove all the clichés associated with the volume.

You can manually delete snapshots (for example, if you need more of)

snapshot data). In addition, the group can delete snapshots automatically in the

following situations:

1 - failure instant free reserve. If the snapshot data consume the snapshot

. reserve, the group either deletes the oldest snapshots to free up space for new

. snapshots or sets the volume and snapshots in offline mode, according to the policy

. you have chosen for the snapshot space recovery.

2 - maximum number of snapshots, created from an agreed timetable. If you set up a

. timetable for the creation of snapshots, you can specify the maximum number of

. photos you want to keep. Once the program creates the maximum number

. clichés, the group will delete the oldest snapshot for planning

. to create a new snapshot.

Snapshot are deleted in the background queue. The group travels

the snapshot page table and decremented the reference count on every shared page.

Any page that has a zero reference count is released into free space. Pages

with a zero reference count are not released because they are shared with

the volume or other snapshots.

Because stereotypes can be automatically deleted, if you want to keep the

given to a snapshot, you can clone the snapshot.

Reserve snapshots use agreement

------------------------------------

The snapshot reserve is consumed only if you write a shared volume or snapshot

page. However, it is difficult to establish a correlation between the amount of data written in one

volume (or snapshot) with the amount of instant reserve consumed as a result.

especially if you have multiple snapshots.

Because the pages consist of several logical blocks, size e/s and distribution

volume merge affect overall performance of e/s in addition to the snapshot

reserve its use.

For example, do much written about a narrow range of logical blocks in a volume

consumes a relatively low amount of reserve of the snapshot. This is because as Scripture

the same logic block more than once, does not require not additional snapshot

reserve.

However, doing random number wrote a range of logical blocks in a

volume can consume a large amount of reserve of the snapshot, because many other

pages are affected.

In general, use instant reserve depends on the following:

1. number of entries that occur in the volume (or snapshot) and at least one

. snapshot exists. In general, more Scriptures tend to use more snapshot reserve.

. Although multiple writes to the same logical block do not require additional

. space.

2 - the range of logical blocks, on which occur the Scriptures. Written in a wide range of

. logical blocks tend to use more instant reserve written in a narrow

. rank, because more of the written word are to different pages.

3. number of snapshots of the volume and timing of the write operations. Most

. snapshots that you create more snapshot reserve is necessary, unless there is

. few entries on the volume or snapshots.

4 - age of snapshot. Snapshots older tend to consume more snapshot reserve only

. the clichés of recent because the group must retain the original data for a

. longer time.

Design Snapshot reserve

-----------------------

You cannot create snapshots until you book instant space. Snapshot reserve

is set as a percentage of the reserve volume (space) for the

volume.

When you create a volume, by default, the volume uses instant group-wide

reserve adjustment. (The reserve in the scale default setting is 100%. You

can change this default value.) You can change the reserve snapshot setting when you

create a volume or later, even if the volume is in use.

Although the snapshot reserve is not used until the volume or writing snapshot occur, it

is consumed immediately free pool space. For example, if you create a

fully allocated (not thin provisioned) 200 GB volume and specify a snapshot of 50%

pool of reserve, free space is immediately reduced to 300 GB (200 GB for the volume

reserve and 100 GB for Snap reserve), even if there are no pages in use.

Therefore, before you create a volume, you should consider how many snapshot

reserves, if any, to assign to the volume. The reserve of the snapshot set to zero (0)

If you do not want to create snapshots.

The optimal size of the snapshot reserve depends on the amount and type of

changes in the volume and the number of shots you want to keep.

By example, if you set the snapshot reserve 100%, and then create a snapshot.

You can write to each byte of the volume without missing snapshot

reserve. However, if you create another snapshot and then write in each byte of

the volume, the first snapshot is deleted in disk space available for the new snapshot. If

you set the instant reserve at 200%, there would be a sufficient reserve of snapshot

for the two snapshots.

A very conservative strategy in terms of instant reserve sizing is to put the snapshot

book value at 100 times the number of shots you want to keep. This

guarantees that keep you at least the number of snapshots, regardless of the

the number of entries on the volume. However, this strategy is generally allocates

book an excessive amount of snapshot, because that rarely crush you all the

the data in a volume during the lifetime of a snapshot.

The best way to instant size reserves is to assign an initial value to the reserve

and watch how instant you can keep over a period of time specified under a

normal workload. If you use tables to create snapshots, allow the

calendar of work for several days.

To get an initial value for a snapshot reserve volume, you must estimate

the quantity and the type of entries in volume and the number of snapshots, you want

keep. For example:

-If you wait a few Scriptures or writings which are concentrated in a narrow range

. logical blocks and you want to keep only a few shots, start with a value

. 30%.

-If you wait several entries or entries that are random in a wide range of

. logical blocks and you want to keep more than a few shots, start with a value

. 100%.

If the snapshots are deleted until you reach the desired number of snapshots,

increase the percentage of snapshot reserve. If you reach the desired number

shots without consuming much of the instant free reserve, decrease the

percentage of reserve snapshot. Continue to follow instant reserve and

adjustments as needed.

How Thin Provisioning button Snapshot Reserve

----------------------------------------------

The snapshot reserve is based on a percentage of volume reserve (allocated space

a volume). For a volume fully provisioned, the reserve volume is equal to the

stated volume size. However, a thin volume put into service, the volume of reserve

is initially very inferior to the reported size (default is subject to minimum volume

10% of the reported size) and increases as the written volume occur.

If you change a thin volume supplied in a volume fully provisioned, the

amount of reserved snapshot increases automatically, because the volume of reserve

increase in the size of the stated volume.

If you change a volume of fully provisioned to thin-provisioned, the amount of

snapshot of reserve decreases automatically, because the volume of reserve declines.

However, if the snapshot resulting reserves will be too small to store all the

existing snapshots, the group will automatically increase the instant reserve

percentage of value that preserves all existing snapshots.

Reducing the use of instant reserve

-------------------------------

Over time, you can reduce the use of instant reserve by periodically (for example,

Once a month) defragmentation of databases and file systems. Defrag operations

try to consolidate segments of files in a volume and, consequently, to reduce the scope

logical blocks addresses used in the pages of the volume.

Defragment the data read operations from one place and then write data to a new

location. So increased use of instant reserve during and immediately

after defragmentation, because existing snapshots will use more of the usual

amount of the snapshot reserve. However, snapshots created after defragmentation

operation must use less instant reserve, because the data on the volume are more contiguous.

When a volume is highly fragmented, potential reduction of the snapshot reserve

use can be dramatic, especially after removing the large before you defragment

snapshots. Defrag operations can also reduce the I/O load on the group,

because the contiguous data makes more efficient i/o operations, in order to improve the

e/s overall performance.

Latest defragmenters are good to reduce the fragmentation that is not seeking to be

too thorough. Some defragmenters also try to combine the inactive data

further restricting the likelihood of changes to shared pages. However, are not

Defragment too frequently.

Sector alignment may also affect the use of snapshot of the space, especially in

larger volumes. File systems must be correctly aligned with the sector. It comes

described in technical reports for VMware and Windows environments.

0000007 a with ERROR_CODE error checking: 0xc000003c

A BSoD make me down, I get two dump files.

It is overtaking such data, can be the memory has been corruption.

I'm trying to find what is happening in the kernel, but in vain.

I check memory and disk use the tools, the two OK.

Here is the information: https://onedrive.live.com/redir?resid=BD718EE48185D4F6! 1042 & authkey =! AHjr7Sp6bQQGAY4 & ithint = the file % 2crar

*******************************************************************************
* *
* Bugcheck analysis *.
* *
*******************************************************************************

Use! analyze - v to obtain detailed debugging information.

Error checking 7 a, {fffff6fc40031ed0, ffffffffc000003c, 3eab3f860, fffff880063daef0}

Page 40ab73 not present in the dump file. Type ".hh dbgerr004" for more details
Page 40bc5c not present in the dump file. Type ".hh dbgerr004" for more details
Probably caused by: fileinfo.sys (fileinfo! FIPfVolumeRundownFilePrefetchStart + 52)

Follow-up: MachineOwner
---------

0: kd >! analyze - v
*******************************************************************************
* *
* Bugcheck analysis *.
* *
*******************************************************************************

KERNEL_DATA_INPAGE_ERROR (7A)
The requested page of kernel data could not be read in. Usually caused by
a bad block in the disk controller or paging file error. See also
KERNEL_STACK_INPAGE_ERROR.
If the error status 0xC000000E, 0xC000009C, 0xC000009D or 0xC0000185,
This means that the disk subsystem has suffered a failure.
If the status of the error is 0xC000009A, it means the query failed because
a file system has failed to move forward.
Arguments:
Arg1: fffff6fc40031ed0, type of lock which took place (1,2,3 or PTE address value)
Arg2: ffffffffc000003c, error state (state code of e/s normally)
Arg3: 00000003eab3f860, the current process (virtual address for the lock type 3 or PTE)
Arg4: fffff880063daef0, virtual address that could not be paged in (or PTE contents if arg1 is a PTE address)

Debugging information:
------------------

Page 40ab73 not present in the dump file. Type ".hh dbgerr004" for more details
Page 40bc5c not present in the dump file. Type ".hh dbgerr004" for more details

ERROR_CODE: (NTSTATUS) 0XC000003C - {}

BUGCHECK_STR: 0x7a_c000003c

DEFAULT_BUCKET_ID: WIN7_DRIVER_FAULT

Nom_processus: eefprocess.exe

CURRENT_IRQL: 0

ANALYSIS_VERSION: 6.3.9600.17029 (debuggers (dbg). 140219-1702) amd64fre

TRAP_FRAME: fffff88008998480-(.trap 0xfffff88008998480)
NOTE: The frame trap does contain not all registers.
Some registry values can be set to zero or incorrect.
Rax = fffff880063d4300 rbx = 0000000000000000 rcx = fffffa8037f4f390
RDX = 0000000000000006 rsi = 0000000000000000 rdi = 0000000000000000
RIP = fffff880063daef0 rsp = fffff88008998618 rbp = 0000000000000000
R8 = 0000000000000000 r9 = 0000000000000000 r10 = fffff88000800008
R11 = fffffa801dc82aa0 r12 = 0000000000000000 r13 = 0000000000000000
R14 = 0000000000000000 r15 = 0000000000000000
iopl = 0 nv in pe of na EI pl nz nc
FileInfo! FIPfOpenAttemptsRundownStart:
fffff880 '063daef0 0000 add byte ptr [rax], al ds:fffff880' 063 d 4300 = b0
Reset the default scope

LAST_CONTROL_TRANSFER: from fffff800018fc438 to fffff80001888f00

STACK_TEXT:
fffff880'089981 has 8 fffff800'018 CF 438: 0000007 a 00000000' fffff6fc '40031ed0 ffffffff' c000003c 00000003' eab3f860: nt! KeBugCheckEx
fffff880 '089981b 0 fffff800' 0187ad8b: fffffa80'37642200 fffff880 '089982e0 fffff800' 01 at 14440 00000000 00000000': nt! : FNODOBFM: 'string' + 0x34cde
fffff880 ' 08998290 fffff800' 018a3da4: 00000000'00000000 00000000'00000008 ffffffff 'ffffffff fffff880' 014e1ab3: nt! MiIssueHardFault + 0x28b
"fffff880 ' 08998320 fffff800' 01886fee: 00000000 ' 00000008 fffffa80 '37f4f390 fffffa80' 1ee60900 fffffa80 ' 1ee60a08: nt! MmAccessFault + 0x11c4
"fffff880 ' 08998480 fffff880' 063daef0: fffff880'063db1ea 00000000 00000000" fffff880 ' 00800008 fffffa80 ' 1e3a3400: nt! KiPageFault + 0x16e
"fffff880 ' 08998618 fffff880' 063db1ea: 00000000 00000000" fffff880 ' 00800008 fffffa80 ' 1e3a3400 00000000'00000801: fileinfo! FIPfOpenAttemptsRundownStart
fffff880 ' 08998620 fffff880' 063dc711: fffffa80'1dc82950 fffffa80 '37f4f390 fffffa80' 37f4f390 00000000' 00000000: fileinfo! FIPfVolumeRundownFilePrefetchStart + 0 x 52
fffff880 '089987b 0 fffff880' 063d07ae: fffffa80'00000001 fffffa80 '1dc82a00 fffffa80' 37f4f390 00000000' 00000000: fileinfo! FIPfFileFSOpPostSetInformation + 0 x 159
"fffff880 ' 08998810 fffff880' 012 a 8242: fffffa80 ' 00000000 00000000 00000000' fffffa80 ' 1ddacb20 00000000' 00000000: fileinfo! FIPostSetInformationCallback + 0 x 236
fffff880 ' 08998890 fffff880' 012a738b: fffffa80 ' 1ddacb20 fffffa80 '1f603d00 fffffa80' 1c60d010 fffffa80 ' 1c60d320: fltmgr! FltpPerformPostCallbacks + 0 x 392
"fffff880 ' 08998960 fffff880' 012a56df: fffffa80 '19486de0 fffffa80' 1ddac810 fffffa80 ' 19486 d 00 fffffa80 ' 1ddac810: fltmgr! FltpLegacyProcessingAfterPreCallbacksCompleted + 0x39b
fffff880 '089989f0 fffff800' 01b6949d: 00000000' 000000a 2 fffff880 ' 08998 c 60 00000000 00000000' fffffa80' 20353170: fltmgr! FltpDispatch + 0xcf
"fffff880'08998 has 50 fffff800'01888153: 00000000 ' 00000150 fffffa80 ' 1c155b60 00000000' 0000007f'000000 0012e928 has 2: nt! NtSetInformationFile + 0 x 909
fffff880' 08998b 70 00000000' 7756012: 00000000'00000000 00000000'00000000 00000000'00000000 00000000'00000000: nt! KiSystemServiceCopyEnd + 0x13
00000000' 0012e908 00000000'00000000: 00000000'00000000 00000000'00000000 00000000'00000000 00000000'00000000: 0x7756012a

STACK_COMMAND: kb

FOLLOWUP_IP:
FileInfo! FIPfVolumeRundownFilePrefetchStart + 52
fffff880'063db1ea 4d8db42488000000 lea r14, [r12 + 88 h]

SYMBOL_STACK_INDEX: 6

SYMBOL_NAME: fileinfo! FIPfVolumeRundownFilePrefetchStart + 52

FOLLOWUP_NAME: MachineOwner

MODULE_NAME: fileinfo

Nom_image: fileinfo.sys

DEBUG_FLR_IMAGE_TIMESTAMP: 4a5bc481

FAILURE_BUCKET_ID: X64_0x7a_c000003c_fileinfo! FIPfVolumeRundownFilePrefetchStart + 52

BUCKET_ID: X64_0x7a_c000003c_fileinfo! FIPfVolumeRundownFilePrefetchStart + 52

ANALYSIS_SOURCE: KM

FAILURE_ID_HASH_STRING: km:x64_0x7a_c000003c_fileinfo! fipfvolumerundownfileprefetchstart + 52

FAILURE_ID_HASH: {2d046ffb-9d55-ace1-dd0b-32387702e9c5}

Follow-up: MachineOwner
---------

0: kd >! fffff6fc40031ed0 pte
GO fffff880063da000
PXE to the EPP of FFFFF6FB7DBEDF88 at FFFFF6FB7DBF1000 at FFFFF6FB7E200188 at FFFFF6FC40031ED0 PTE POE
contains 000000040CE04863 contains 000000040CE03863 contains 00000003E0319863 contains 00000003EAB3F860
NFP 40ce04 - DA - KWEV NPF 40ce03 - DA - KWEV NPF 3e0319 - DA - invalid KWEV
Transition: 3eab3f
Protect: 3 - Executionlecture

0: kd >! 3eab3f NFP
NFP 003EAB3F at the address FFFFFA800BC01BD0
Flink FFFFFA8037642220 blink / hand count 00000000 pteaddress FFFFF6FC40031ED0
stock count 0001 used reference count 0000 Cached color priority 0 0
catering pte C1DF00000060 containing the R page watch 3E0319
ReadInProgress

0: kd >! treat 0-1
PROCESS fffffa801b96b060
SessionId: 0 Cid: 0704 Peb: 7fffffd3000 ParentCid: 072 c
DirBase: 639e7000 ObjectTable: fffff8a010225ac0 HandleCount: 84.
Image: eefprocess.exe

0: kd >! thread 0-1
WIRE fffffa801c155b60 Teb 0704.14d4 Cid: 000007fffffde000 Win32Thread: fffff900c073e010 PROCESSOR 0
0: kd >! thread fffffa801c155b60
WIRE fffffa801c155b60 Teb 0704.14d4 Cid: 000007fffffde000 Win32Thread: fffff900c073e010 PROCESSOR 0
The IRP list:
fffffa801ddac810: (0006, 03e8) flags: 00060830 Mdl: 00000000
Borrow not identity
DeviceMap fffff8a000006110
Owning process fffffa801b96b060 Image: eefprocess.exe
Joint process s/o Image: n/a
Wait start TickCount 6381243 ticks: 2 (0:00:00:00.031)
Context switch count 287 IdealProcessor: 7 LargeStack
Get_user_time 00:00:00.015
KernelTime 00:00:00.046
0x00000001400023a8 Win32 start address
Stack Init fffff88008998d70 current fffff88008997f00
Basic fffff88008999000 limit fffff88008991000 dial 0
Priority 8 BasePriority 8 UnusualBoost 0 ForegroundBoost 0 IoPriority 2 PagePriority 5
Child-SP RetAddr: Args to child: call Site
fffff880'089981 has 8 fffff800'018 CF 438: 0000007 a 00000000' fffff6fc '40031ed0 ffffffff' c000003c 00000003' eab3f860: nt! KeBugCheckEx
fffff880 '089981b 0 fffff800' 0187ad8b: fffffa80'37642200 fffff880 '089982e0 fffff800' 01 at 14440 00000000 00000000': nt! : FNODOBFM: 'string' + 0x34cde
fffff880 ' 08998290 fffff800' 018a3da4: 00000000'00000000 00000000'00000008 ffffffff 'ffffffff fffff880' 014e1ab3: nt! MiIssueHardFault + 0x28b
"fffff880 ' 08998320 fffff800' 01886fee: 00000000 ' 00000008 fffffa80 '37f4f390 fffffa80' 1ee60900 fffffa80 ' 1ee60a08: nt! MmAccessFault + 0x11c4
"fffff880 ' 08998480 fffff880' 063daef0: fffff880'063db1ea 00000000 00000000" fffff880 ' 00800008 fffffa80 ' 1e3a3400: nt! KiPageFault + 0x16e (TrapFrame @ fffff880'08998480)
"fffff880 ' 08998618 fffff880' 063db1ea: 00000000 00000000" fffff880 ' 00800008 fffffa80 ' 1e3a3400 00000000'00000801: fileinfo! FIPfOpenAttemptsRundownStart
fffff880 ' 08998620 fffff880' 063dc711: fffffa80'1dc82950 fffffa80 '37f4f390 fffffa80' 37f4f390 00000000' 00000000: fileinfo! FIPfVolumeRundownFilePrefetchStart + 0 x 52
fffff880 '089987b 0 fffff880' 063d07ae: fffffa80'00000001 fffffa80 '1dc82a00 fffffa80' 37f4f390 00000000' 00000000: fileinfo! FIPfFileFSOpPostSetInformation + 0 x 159
"fffff880 ' 08998810 fffff880' 012 a 8242: fffffa80 ' 00000000 00000000 00000000' fffffa80 ' 1ddacb20 00000000' 00000000: fileinfo! FIPostSetInformationCallback + 0 x 236
fffff880 ' 08998890 fffff880' 012a738b: fffffa80 ' 1ddacb20 fffffa80 '1f603d00 fffffa80' 1c60d010 fffffa80 ' 1c60d320: fltmgr! FltpPerformPostCallbacks + 0 x 392
"fffff880 ' 08998960 fffff880' 012a56df: fffffa80 '19486de0 fffffa80' 1ddac810 fffffa80 ' 19486 d 00 fffffa80 ' 1ddac810: fltmgr! FltpLegacyProcessingAfterPreCallbacksCompleted + 0x39b
fffff880 '089989f0 fffff800' 01b6949d: 00000000' 000000a 2 fffff880 ' 08998 c 60 00000000 00000000' fffffa80' 20353170: fltmgr! FltpDispatch + 0xcf
"fffff880'08998 has 50 fffff800'01888153: 00000000 ' 00000150 fffffa80 ' 1c155b60 00000000' 0000007f'000000 0012e928 has 2: nt! NtSetInformationFile + 0 x 909
fffff880' 08998b 70 00000000' 7756012: 00000000'00000000 00000000'00000000 00000000'00000000 00000000'00000000: nt! KiSystemServiceCopyEnd + 0 x 13 (TrapFrame @ fffff880'08998be0)
00000000' 0012e908 00000000'00000000: 00000000'00000000 00000000'00000000 00000000'00000000 00000000'00000000: 0x7756012a

0: kd >! fffffa801ddac810 of the IORP
IRP is active with 9 battery 9 is current (= 0xfffffa801ddacb20)
No Mdl: The system buffer = fffffa801b6db460: thread fffffa801c155b60: Irp stack trace.
cmd flg cl device file completion-context
[0, 0] 0 2 00000000 00000000 00000000 - 00000000

Args: 00000000 00000000 00000000 ffffffffc000003c
[0, 0] 0 0 00000000 00000000 00000000 - 00000000

Args: 00000000 00000000 00000000 00000000
[0, 0] 0 0 00000000 00000000 00000000 - 00000000

Args: 00000000 00000000 00000000 00000000
[0, 0] 0 0 00000000 00000000 00000000 - 00000000

Args: 00000000 00000000 00000000 00000000
[0, 0] 0 0 00000000 00000000 00000000 - 00000000

Args: 00000000 00000000 00000000 00000000
[0, 0] 0 0 00000000 00000000 00000000 - 00000000

Args: 00000000 00000000 00000000 00000000
[0, 0] 0 0 00000000 00000000 00000000 - 00000000

Args: 00000000 00000000 00000000 00000000
[6: 0] 0 2 fffffa801949e030 00000000 fffff880012a8980-fffffa801f603ab0
\FileSystem\Ntfs fltmgr! FltpSynchronizedOperationCompletion
Args: 00000000 00000000 00000000 00000001
> [6: 0] 0 0 fffffa8019486de0 fffffa801b577810 00000000-00000000
\FileSystem\FltMgr
Args: 000000 has 2 0000000 has fffffa8020353170 00000001

0: kd > dt _FILE_OBJECT fffffa8020353170

Help check

ALSO

The requested page of kernel data could not be read in. Usually caused by
a bad block in the file paging or the disk controller error. See also
KERNEL_STACK_INPAGE_ERROR.

Bluescreen error 0xc0000018 on Windows 8

* Display for the customer account *.

The PC is an Asus N82JB with 64-bit Windows 8 and meets the minimum requirements of the system. Cold start the PC, the PC has frozen, then performed a hard reset and the error on blue screen appeared read 0xc0000018error. The client is trying a system restore.

This could be a driver problem? I find no information about this error, except the one that applies to the office, so is therefore irrelevant in the present case. Can anyone provide more information about this error?

Thank you!

Hello

Definitely check with the support of ASUStek, their documentation online and drivers, and
Ask problems known in their forums. Given that Windows 8 is very recent, you should
check closely with them in the coming weeks or more. Look for the updates of the
BIOS, drivers low-level chipset and drivers of devices shipped major. Also check with
3rd manufacturers part such as video, sound and network for updated drivers.

ASUStek support
http://support.ASUS.com/

ASUStek Forums
http://VIP.ASUS.com/forum/default.aspx?SLanguage=en-us

ASUStek drivers
http://support.ASUS.com/download/download.aspx?SLanguage=en-us

========================================================

Note There are not specific online and/or resources KB for Windows 8 yet if you
may need to translate some of the Vista and Windows 7 methods. Special
Note must use Win Key + X and Win Key + W (to access control panel, run,
and settings according to needs). Win + D key calls the office and using the key to victory
Activates / deactivates the desktop and start screen.

Translate these methods of Windows 8 (there is little real difference):

Resolve errors stop (blue screen) in Windows 7 - has a section for if you can or cannot start Windows.
http://Windows.Microsoft.com/en-us/Windows7/resolving-stop-blue-screen-errors-in-Windows-7

Check this thread for more information using BlueScreenView, MyEventViewer and other methods
to troubleshoot the BlueScreens - top 3 responses (+ 1 other).

http://answers.Microsoft.com/en-us/Windows/Forum/Windows_7-system/sometimes-i-get-a-blue-screen-when-using-IE-8/c675b7b8-795f-474d-a1c4-6b77b3fcd990

We can analyze the minidumps if make you it available to the SkyDrive or another file
sharing sites (such as MediaFire). If you have problems to download the copy of minidumps
for the office or in the Documents folder and download them from there.

ZIP or download the content of the C:\Windows\minidump

Use SkyDrive to upload collected files
http://social.technet.Microsoft.com/forums/en-us/w7itproui/thread/4fc10639-02dB-4665-993a-08d865088d65

-------------------------------------------------------------------------

Also this, so you can see the probable bluescreens.

Windows Vista restarts automatically if your computer encounters an error that requires him to plant.
(also Windows 7 and 8)
http://www.winvistatips.com/disable-automatic-restart-T84.html

===========================================================

BCCode: 18 0 x 00000018

This majority is often a driver problem so try to update the main drivers, BIOS and low level
chipset drivers.

Other possibilities include malware and antivirus/antispyware/security programs even and
others. See the driver update methods in the following message when you get to this part of the
convenience store and return to the if necessary troubleshooting tool.

BCCode: 18 0 x 00000018

Bug Check 0x18: REFERENCE_BY_POINTER

Bug REFERENCE_BY_POINTER control has a value of 0 x 00000018. This indicates that the reference of an object count is illegal for the current state of the object.

Important info if you received a STOP Code

If you received a blue error screen, or a stop code, the computer stopped abruptly to protect against data loss. A hardware device, its driver or software might have caused this error. If your copy of Windows is shipped with your computer, contact your computer manufacturer. If you purchased Windows separately from your computer, Microsoft provides support. To find contact information for Microsoft or the manufacturer of your computer, Contact Support.

If you have experience with computers and try to recover from this error, follow the steps in the Microsoft article solution STOP (blue screen) errors in Windows.

These actions could prevent a mistake like this does not happen:

Download and install updates for your computer from Windows Update device drivers.
Scan your computer for computer viruses.
Check your hard drive for errors.

REFERENCE_BY_POINTER settings

The following settings are displayed on the blue screen.

Parameter	Description
1	Object type of the object whose reference count is lowering.
2	Object whose reference count is lowering.
3	Reserved
4	Reserved

Cause

The reference of an object count is illegal for the current state of the object. Whenever a driver uses a pointer to an object, the driver calls a kernel routine to increase the number of references to the subject of a. When the driver is done with the pointer, the driver calls another routine of core to reduce the number of references by one.

Pilots must match calls to routines that increase (reference) and decrease (dereferencing) the reference count. This bug check is caused by inconsistency in the reference of the object count. As a general rule, the inconsistency is caused by a driver that reduces the number of references of an object calls too many times, additional dereference the object. This bug check can occur because the object reference count tends to zero, although there are still open handles to the object. It might also occur when the object reference count is less than zero, if there are handles open to the object.

Resolution

Make sure that the driver matches calls to routines that increase and decrease the number of references to the object. Make sure that your driver does no additional calls to routines that dereferencing the object (see parameter 2).

You can use a debugger to help analyze this problem. To find the handle and pointer to rely on the object, use it ! object debugger command.

KD >! address of object

Where address is the address of the object given in parameter 2.

BCCode: 18 0 x 00000018<-- read="" this="">
http://www.faultwire.com/solutions-fatal_error/reference-by-pointer-0x00000018-* 1039.html? = date order

-------------------------------------------------------------------------

Look in the Event Viewer to see if something is reported on those.
http://www.computerperformance.co.UK/Vista/vista_event_viewer.htm

MyEventViewer - free - a simple alternative in the standard Windows Event Viewer.
TIP - Options - Advanced filter allows you to see a period of time instead of the entire file.
http://www.NirSoft.NET/utils/my_event_viewer.html

I hope this helps.

Rob Brown - Microsoft MVP<- profile="" -="" windows="" expert="" -="" consumer="" :="" bicycle=""><- mark="" twain="" said="" it="">

ESX3.5 U2 blue screen with processor AMD daul

We have a problem with ESX3.5 u2 start with the daul core AMD CPU, when ESX3.5 Initializes the timer, it willreport the error as follows:
;----
TSC: cpu0:0 4552357591) initialization calendar...
TSC: cpu0:0 10150276678) Init: VMKernel:718: cpu 0: tsc measured speed 2800112697 Hz, 2800112697 Hz (using TSC) CPU speed, bus speed 200008050 Hz
Could not find a usable reference clock
;----
and after Debug us we found, it seems, it cannot initialize the ACPI PM timer and daul AMD processor cannot find the reference count, so it will signal you the error.and qual hearts AMD processor, is not also initialize the ACPI PM timer, but he will refer to the timer with TSC. So, how ESX3.5 u2 detect ACPI PM timer and initialize it? By the FACT of the ACPI table? How ESX3.5 U2 can do qual AMD processor can reference with TSC, why Daul CPU can not?
and we have another issue with ESX3.5 u2 start when we activated the HPET function, it will report as follows;
;----
init IO_APIC IRQ
IO-APIC (apicid-PIN) 2-2, 2-16, 2-17, 2-18, 19 / 2, 2-20, 2-21, 2-22, 2-23 not connected.
.. TIMER: vector = 0 x 31 pin1 = 0 pin 2 =-1
.. MP-BIOS bug: 8254 timer connected to IO-APIC
... try to put in place the timer (IRQ0) through the 8259... has failed.
... try to set up the timer like Virtual Wire IRQ... failed.
... try to set up the timer like ExtINT IRQ... unexpected trap IRQ vector 07
trap unexpected IRQ to the vector 07
failure: (.)
Kernel panic: IO-APIC + timer does not work! rant [email protected]
;----
So, how ESX3.5 U2 to check with these?
Thank you very much.

Is what brand/model of server? Is it in the version of ESX 3.5 certified list? What family of AMD processor?

I got something similar on a whitebox server before where a bunch of stuff from power management enabled (' Quick ' n and calm "or something like that) which distorted speed CPU and timers like her in tray autour CPU running just as fast as necessary. For virtualization, you need all this out...

Memory management tips?

Hello Experts.
I'm experimenting with dynamically created and destroyed with the ancestors script objects (strange how the child object seems to create the instance of the ancestor?). Some of them create and use objects image during their lifetime, referenced by the properties own scripts. My main question is if I put the reference to the script object EMPTY or no image objects created by the garbage collector, so script?
Or do I have to create a certain custom on ~ destroy() style handlers for each script and remember zero each image object /ancestor, which creates the instance of script?
That popular memory in general management advice!
Thank you
Richard

As long as the only reference to the picture object is a property of a script object, as soon as the script object is garbage collected (its reference count drops to zero) the object of the image will be deleted with and his freed memory. However, there is nothing wrong with playing it safe and to be part of yourself, like your theoretical "on mDispose()" Manager.

Integration Siebel cat > cca813
Hi, I imported the multichannelC.def driver (language: PTB) in Siebel and configured agents, but when I login, it load the toolbar and the message "disconnected".

Could you help me please?

TKS

Journal of the tw - summit_xxxxx.log

CustomId: 352003
: INFO: Opening [2009-10-28 14:36:24:625] Session of Application ADMIN succeed!
[14:36:24:625 10/28/2009]: INFO: from GetAllMsgFunc wire succeed!
[14:36:24:625 10/28/2009]: INFO: from ParseSOAPFunc wire succeed!
[14:36:24:625 10/28/2009]: DEBUG: from GetAllMsgFunc wire
[14:36:24:625 10/28/2009]: DEBUG: from ParseSOAPFunc wire
[14:36:24:625 10/28/2009]: DEBUG: from ProcessEventFunc wire
[2009-10-28 14:36:24:625]: DEBUG:Begin: ProcessEventFunc
[2009-10-28 14:36:24:625]: DEBUG:End: ProcessEventFunc
[14:36:24:625 10/28/2009]: INFO: from ProcessEventFunc wire succeed!
Initialization of: INFO: Driver [2009-10-28 14:36:24:625] succeed!
[14:36:24:625 10/28/2009]: INFO: Add the driver instance in the list of drivers.
: INFO: Driver [2009-10-28 14:36:24:625] increases the reference count = 1
[14:36:24:640 10/28/2009]: DEBUG: add new service in the service list.
[2009-10-28 14:36:25:687]: ERROR:ParseSOAPFunc: Erro CCA! Empresa inatingivel. (* CCA error! Company unattainable.*)
: DEBUG:ParseSOAPFunc [14:36:25:687 10/28/2009]: failure of count = 1
[2009-10-28 14:36:26:718]: ERROR:ParseSOAPFunc: Erro CCA! Empresa inatingivel.
: DEBUG:ParseSOAPFunc [14:36:26:718 10/28/2009]: failure of count = 2
[2009-10-28 14:36:27:734]: ERROR:ParseSOAPFunc: Erro CCA! Empresa inatingivel.
: DEBUG:ParseSOAPFunc [14:36:27:734 10/28/2009]: failure of count = 3
[2009-10-28 14:36:28:765]: ERROR:ParseSOAPFunc: Erro CCA! Empresa inatingivel.
: DEBUG:ParseSOAPFunc [14:36:28:765 10/28/2009]: failure of count = 4
[2009-10-28 14:36:29:781]: ERROR:ParseSOAPFunc: Erro CCA! Empresa inatingivel.
: DEBUG:ParseSOAPFunc [14:36:29:781 10/28/2009]: failure of count = 5
[2009-10-28 14:36:30:828]: ERROR:ParseSOAPFunc: Erro CCA! Empresa inatingivel.
: DEBUG:ParseSOAPFunc [14:36:30:828 10/28/2009]: failure of count = 6
[2009-10-28 14:36:31:828]: ERROR:ParseSOAPFunc: Erro CCA! Empresa inatingivel.
: DEBUG:ParseSOAPFunc [14:36:31:843 10/28/2009]: failure of count = 7
[2009-10-28 14:36:32:875]: ERROR:ParseSOAPFunc: Erro CCA! Empresa inatingivel.
: DEBUG:ParseSOAPFunc [14:36:32:875 10/28/2009]: failure of County = 8
[2009-10-28 14:36:33:875]: ERROR:ParseSOAPFunc: Erro CCA! Empresa inatingivel.
: DEBUG:ParseSOAPFunc [14:36:33:875 10/28/2009]: count failure = 9
[2009-10-28 14:36:34:937]: ERROR:ParseSOAPFunc: Erro CCA! Empresa inatingivel.
: DEBUG:ParseSOAPFunc [14:36:34:937 10/28/2009]: failure of count = 10
[2009-10-28 14:36:34:937]: ERROR:No backup CCA configured server, could not perform switching!
[14:36:34:937 10/28/2009]: FATAL: ParseSOAPFunc: stop!
: INFO: LogoutAgent [2009-10-28 14:36:34:937]: fatal error! Servidor CCA desconectado! Logout automático!
[2009-10-28 14:36:39:984]: INFO: Stopping GetAllMsg wire succeed!
[14:36:40:078 10/28/2009]: DEBUG: output wire GetAllMsgFunc
[2009-10-28 14:36:44:984]: INFO: Stopping ParseSOAP wire succeed!
: INFO: Stopping [2009-10-28 14:36:49:984] ProcessEvent Thread succeed!
[14:36:49:984 10/28/2009]: DEBUG: output wire ParseSOAPFunc
[14:36:50:062 10/28/2009]: DEBUG: output wire ProcessEventFunc
[2009-10-28 14:38:30:875]: DEBUG:Remove list of service service.
[2009-10-28 14:38:30:875]: INFO: Driver placed reference, count = 0
Driver instance of: INFO: Remove [2009-10-28 14:38:30:875] from the list of drivers.
Connection: INFO: Max [2009-10-28 14:58:05:703]: 2
: INFO: GetUniqueID [2009-10-28 14:58:05:703]
[2009-10-28 14:58:05:703]: DEBUG:ProcessId: 3520, seeds: 4
Parameter [2009-10-28 14:58:05:703]: DEBUG:TaW Driver:
URL = http://172.23.2.12:8080/CCA/services /.
AutoProxyRefreshInterval = - 1
LogSOAPFile = twsoap - summit.log
BackupURL =
EventRetry = 10
UnknownCCAValue =?
RefreshUserExtensionInterval = 3600
LogFile = tw - summit.log
SOAPRetry = 3
OnHookAtLogout = TRUE
ProxyLocation =
CountryCodeDelimiter =.
DefaultCountryCode = 1
ProxySupport = disabled
MaxMessageNumber = 50
MessageInterval = 1000
MaxLogKB = 90240
AdminUserName = admin
CompanyName = system
MaxAgentNumber = - 1
ReleaseLogHandle = TRUE
VoiceDescription = [ANI]
EnableSOAPTrace = FALSE
LogDebug = TRUE
AdminPassword = admin
You need to configure your communication settings driver to match your settings of CCA as an alias of the company. The error is saying that the company is not found. Failure in the DEF file is top of society which is probably not in your CCA.

Summary of complex scenario (counties)

Hi all

Looking for inspiration and ideas with this little teaser

I'm trying to find a common approach to the production of four counts (A, B, C and D) indicated in lines 1 to 5 below. This function of a table that contains about 4 000 entries.

In affect, I try to summarize the different permutations in the classifications of four cells show in the image above, using a common approach. I might add a column for each company and use wild-card account, but I'd rather keep the table of (imported) origin of tact - I know that I could build a table of distinct society in another piece of work, but there must be a smarter way to achieve this.

The legend below presents an overview of how the data looks like with different options (style truth table). The reference column contains a master list starting with RE - unique references.

NC and P columns contain company names. There are two additional columns and the television that contain one or more numbers starting with- and TV - respectively.

The idea is to check that the RE has a link to references or TV in the respective columns when there is a company name in the naming context and / column P

There are a number of scenarios based on conditional counts in two columns: following NC and P with the possible permutations:

Swap 1: Where there is a company name in the column "NC" count the number of:

RE links to OF
RE TV links
RE links to OF and TV
RE completely unrelated
More than one company with no link

Swap 2: Where there is a company name in the column 'P' count the number of:

RE links to OF
RE TV links
RE links to OF and TV
RE completely unrelated
More than one company with no link

Switch 3: When there is an entry in the column "NC" both "P":

RE links to OF
RE TV links
RE links to OF and TV
RE completely unrelated
More than one company with no link

At this point, my head began to ache - would be very grateful and ideas on how this could be done.

Thanks in advance

Perry

If I understand correctly, you can compile your different counts using COUNTIF with the * wildcard, something like this:

For more convenience in formulas to set up your data in a table and your account in another.

Here are the formulas in the table 'Account':

B2:	= COUNTIFS(Data::$B,"RE",Data::C,"<>"&"",Data::$F,"DE")
C2:	= COUNTIFS(Data::$B,"RE",Data::D,"<>"&"",Data::$F,"DE")
D2:	= SUM(B2:C2)
B3:	= COUNTIFS(Data::$B,"RE",Data::C,"<>"&"",Data::$G,"TV")
C3:	= COUNTIFS(Data::$B,"RE",Data::D,"<>"&"",Data::$G,"TV")
D3:	= SUM(B3:C3)
B4:	= COUNTIFS(Data::$B,"RE",Data::C,"<>"&"",Data::$F,"DE",Data::$G,"TV")
C4:	= COUNTIFS(Data::$B,"RE",Data::D,"<>"&"",Data::$F,"DE",Data::$G,"TV")
D4:	= SUM(B4:C4)
B5:	= COUNTIFS(Data::$B,"RE",Data::C,"<>"&"",Data::$F,"",Data::$G,"")
C5:	= COUNTIFS(Data::$B,"RE",Data::D,"<>"&"",Data::$F,"",Data::$G,"")
D5:	= SUM(B5:C5)

You can learn more about COUNTIF here (you can also access it via help > formulas & functions help menu.)

For more information on the search for wildcards of 'wild' in help > formulas & functions help

The function works with pairs of columns-condition, where you add a reference to a beach (usually a column because it's more convenient) and then apply a 'condition' against it, indeed the application of several filters.

You can add pairs more than necessary.

use the contents of the cell to a cell reference of formula calculate

In Apple "Numbers", in order to update a spreadsheet each time that the values are added, I first need to calculate the number of days between the date of first entry and last date, using the COUNTA() method. Gives me the line number of the last entry. For this I "CONCATENATE" a column letter known to a cell reference.

Using this cell reference, I would like to know how can I use the contents of the cell to update the information contained in the rest of the worksheet. Is this possible?

Any help would be appreciated.

Thanks in advance.

I have a problem following exactly what you want to do, but think that INDIRECT can help. More here.

SG

Q reference count

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