Unable to understand the output...

Hi gurus

I was reading a little code and unable to understand a single line, can someone guide me to understand this line.

Code example

Set serveroutput on

DECLARE

n number;

BEGIN

n: = 2 * 33.

dbms_output.put_line (n);

end;

Result

8589934592


Now, I would like to know how the calculation performed for n: = 2 * 33 in the background so it become 8589934592 result. Thanks in advance.

One of these curiosities PL/SQL in SQL vs. In PL/SQL * is same as it is built based on POWER. 2 * 33 is identical to POWER (2.33). It is documented in the Precedence of the operators:

Table 2-3. precedence of operators

Operator Operation

**

Exponentiation

SY.

Tags: Database

Similar Questions

  • Unable to locate the output interface...

    Hello

    I configured a VPN of Lan Lan 2 and it works fine.

    VPN use IKEv2 authentication and certificate.

    Computer 1 can reach 2 computer without problem.

    Computer 1, I tried to access the intellectual property inside 2 (ping, AMPS...) but I get this error: unable to locate the output interface

    I do not understand why I can access computer IP 2 but not intellectual property inside 2 people 2 IPs are on the same network and packets go through the same features...

    How can I solve this problem?

    Thanks for your help,

    Patrick

    Hello

    I imagine that the two devices we're talking about are firewalls Cisco? I mean the VPN devices.

    Firewalls Cisco do not allow ICMP behind an interface to another interface on the same device. The only exception is when traffic is coming through VPN and enter a configuration command specific to the device that you're trying to ICMP behind a VPN connection.

    So if 1 computer needs ICMP inside 2 then the firewall with 2 inside interface must be configured with the command

    access to the administration

    Same configuration is required on the other firewall if 2 computer needs ICMP inside 1

    It can also be NAT associated configurations that may require changes, but it depends on the level of your firewall software which we don't know.

    -Jouni

  • Unable to understand the result of the loop


    Hi all

    I got the tracking code:

    Code

    Set serveroutput ON

    DECLARE
    TYPE IS VARRAY months_array (12) STRING (3 CHAR);
    MONTHS months_array: = months_array ('Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec');
    BEGIN
    < < list_month > >
    FOR month number IN REVERSE 6.12
    LOOP
    dbms_output.put_line (' number month = ' | number month |) "Items =' | MONTH (MONTHNO));
    CONTINUE list_month
    WHEN number month < 9;
    END LOOP;
    END;

    Code output

    number months = 12 = Dec

    number month = 11 = Nov

    number months = 10 items = Oct

    number month = 9 elements = Sep

    number month = 8 = August

    number month = 7 cells = Jul

    number months = 6 items = Jun

    My question is if you notice the code condition which is CONTINUE list_month number months WHEN < 9 means loop must be output if number month < 9 and according to code loop starting with number months = 12 so system should display the 1st line of output that is number months = 12 elements = Dec , then exit because the loop exit condition is true that is list_month MORE WHEN number month < 9. Please, help me to understand the output.

    Hi Muzz,

    try something like this

    BEGIN
      FOR monthno IN REVERSE 6..12
      LOOP
        exit when monthno < 9;
        dbms_output.put_line('monthno= '||monthno||' Elements='||MONTH(monthno));
      END LOOP;
    END;
    

    concerning

    Kay

  • Unable to display the output of a concurrent program

    Dear members,

    I use Oracle R12.1.3 Vision Instance.

    I ran the program create accounting. When I click on the button to display output, the browser gets displayed and gets immediately closed. I'm unable to view the output.

    Can someone help me please to solve this issue?

    Thanks in advance.

    Kind regards.

    I faced this problem once. That's the problem with the settings of Internet Explorer.
    Check in the settings of IE-> tools - > Internet Options-> Security tab-> the zone (Intranet, for the most part)-> enable downloads.

    By
    VAMSi

  • Unable to display the output in the next block

    Hello
    I tried the next block... But unable to see the output
    Could you suggest me in this...
    0-----)
    
    PL/SQL procedure successfully completed.
    Although made setserverout

    but of no use

    Please help me

    Published by: josh1612 on April 5, 2010 12:14 AM

    Hello

    Check if the script given below will print something...

    DECLARE
      ..
    BEGIN
       DBMS_OUTPUT.PUT_LINE('START OF ANONYMOUS BLOCK');
      ..
       DBMS_OUTPUT.PUT_LINE('END OF ANONYMOUS BLOCK');
    END;
    

    If so, check your logic for the results expected, if not... Check the settings of sqlplus

    * 009 *.

    Edited to take off your real script :) forums

    Published by: 009 April 5, 2010 12:18 AM

  • Unable to understand the code output

    Hi all

    I was reading one of the codes and unable to understand its outcome, I'm really grateful if someone guide me to understand. Thanks in advance.

    Code example

    create or replace package Mp_pkg

    IS

    P_nn TYPE TABLE IS NUMBER

    INDEX BY VARCHAR2 (10);

    END;

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

    DECLARE

    l_numbers Mp_pkg.p_nn;

    Start

    FOR indx IN 1... 50

    loop

    (indx) l_numbers: = indx;

    END loop;

    dbms_output.put_line (l_numbers. (Last);

    END;

    ----------

    Result

    9

    -----------

    According to my understanding, result should be 50 because the last value of l_number is 50...

    Please guide...


    Hello

    As John said, the VARCHAR2 '9 ' (or any string '9') is longer than the string of 50', for the same reason that the string 'Z' is longer than the string "THE".

    What do we mean when we say this string x is greater than string y?  We mean that

    1. The 1 x chraracter comes after the 1st character of y (in sort order), or, if these characters are equal
    2. x has a 2nd character, but is not, or, if both are more than 1 character
    3. The substring SUBSTR (x, 2) > SUBSTR (y, 2)

    What (if any) strings mean is not serious, so

    • '9' is greater than '50',
    • 'zero' is greater than 'a '.
    • 'small' is greater than 'big '.
    • 'MIN' is greater than 'MAX '.
    • 'yesterday' is greater than 'tomorrow', and
    • "1 January 1900' is superior to" December 31, 2015 ".

    According to rule 1 Correction: rule 3 is the reason why 'MIN' is greater than 'MAX '; the others are all due to rule 1

  • Understanding the output - AC db_stat, lockers, locks and lock objects

    I would like to understand this output. 80000045 transaction holds a lock for writing on page 11 and is waiting for a lock for writing on page 13, which, however, is not granted as long as transaction 80000051 retains its read lock.

    Is there something in this table explaining why 80000051 doesn't seem unwilling to move and release the read lock?

    And what makes SD = 37 average?

    Michael Ludwig
    Locks grouped by lockers:
    Locker   Mode      Count Status  ----------------- Object ---------------
          a1 dd=37 locks held 1    write locks 0    pid/thread 1284/4896
          a1 READ          1 HELD    test.dbxml                handle        2
          a4 dd=36 locks held 0    write locks 0    pid/thread 1284/4896
          a5 dd=35 locks held 1    write locks 0    pid/thread 1284/4896
          a5 READ          1 HELD    test.dbxml                handle        4
          a8 dd=34 locks held 0    write locks 0    pid/thread 1284/4896
          a9 dd=33 locks held 1    write locks 0    pid/thread 1284/4896
          a9 READ          1 HELD    test.dbxml                handle        6
          ac dd=32 locks held 0    write locks 0    pid/thread 1284/4896
          ad dd=31 locks held 1    write locks 0    pid/thread 1284/4896
          ad READ          1 HELD    test.dbxml                handle        8
          b0 dd=30 locks held 0    write locks 0    pid/thread 1284/4896
          b1 dd=29 locks held 1    write locks 0    pid/thread 1284/4896
          b1 READ          1 HELD    test.dbxml                handle       10
          b4 dd=28 locks held 0    write locks 0    pid/thread 1284/4896
          b5 dd=27 locks held 2    write locks 0    pid/thread 1284/4896
          b5 READ          1 HELD    test.dbxml                handle       12
          b5 READ          6 HELD    test.dbxml                handle        0
          b8 dd=26 locks held 0    write locks 0    pid/thread 1284/4896
          bc dd=25 locks held 1    write locks 0    pid/thread 1284/4896
          bc READ          1 HELD    test.dbxml                handle       14
          bf dd=24 locks held 0    write locks 0    pid/thread 1284/4896
          c0 dd=23 locks held 2    write locks 0    pid/thread 1284/4896
          c0 READ          1 HELD    test.dbxml                handle       16
          c0 READ          2 HELD    test.dbxml                handle        0
          c3 dd=22 locks held 0    write locks 0    pid/thread 1284/4896
          f1 dd=21 locks held 2    write locks 0    pid/thread 1284/4896
          f1 READ          1 HELD    test.dbxml                handle       18
          f1 READ          1 HELD    test.dbxml                handle        0
          f4 dd=20 locks held 0    write locks 0    pid/thread 1284/4896
          f5 dd=19 locks held 0    write locks 0    pid/thread 1284/2648
          f6 dd=18 locks held 0    write locks 0    pid/thread 1284/2648
          f7 dd=17 locks held 0    write locks 0    pid/thread 1284/2648
          f8 dd=16 locks held 0    write locks 0    pid/thread 1284/2648
          f9 dd=15 locks held 0    write locks 0    pid/thread 1284/2648
          fa dd=14 locks held 0    write locks 0    pid/thread 1284/2648
          fb dd=13 locks held 0    write locks 0    pid/thread 1284/2648
          fc dd=12 locks held 0    write locks 0    pid/thread 1284/4184
          fd dd=11 locks held 0    write locks 0    pid/thread 1284/2672
          fe dd=10 locks held 0    write locks 0    pid/thread 1284/4184
          ff dd= 8 locks held 0    write locks 0    pid/thread 1284/2672
         100 dd= 7 locks held 0    write locks 0    pid/thread 1284/2672
         101 dd= 5 locks held 0    write locks 0    pid/thread 1284/2648
         102 dd= 4 locks held 0    write locks 0    pid/thread 1284/2648
    80000044 dd= 9 locks held 0    write locks 0    pid/thread 1284/4184
    80000045 dd= 9 locks held 1    write locks 1    pid/thread 1284/4184
    80000045 WRITE         1 WAIT    test.dbxml                page         13
    80000045 WRITE         1 HELD    test.dbxml                page         11
    8000004d dd= 6 locks held 0    write locks 0    pid/thread 1284/4100
    8000004e dd= 6 locks held 0    write locks 0    pid/thread 1284/4100
    8000004e READ          1 WAIT    test.dbxml                page         11
    80000051 dd= 3 locks held 1    write locks 0    pid/thread 1284/2648
    80000051 READ          1 HELD    test.dbxml                page         13
    80000054 dd= 2 locks held 0    write locks 0    pid/thread 1284/5816
    80000054 READ          1 WAIT    test.dbxml                page         13
    80000055 dd= 1 locks held 0    write locks 0    pid/thread 1284/4824
    80000055 READ          1 WAIT    test.dbxml                page         13
    80000056 dd= 0 locks held 0    write locks 0    pid/thread 1284/2648
    80000057 dd= 0 locks held 0    write locks 0    pid/thread 1284/2648
    80000057 READ          1 WAIT    test.dbxml                page         11
    =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
    Locks grouped by object:
    Locker   Mode      Count Status  ----------------- Object ---------------
    80000045 WRITE         1 HELD    test.dbxml                page         11
    8000004e READ          1 WAIT    test.dbxml                page         11
    80000057 READ          1 WAIT    test.dbxml                page         11
    
          b1 READ          1 HELD    test.dbxml                handle       10
    
          ad READ          1 HELD    test.dbxml                handle        8
    
          bc READ          1 HELD    test.dbxml                handle       14
    
    80000051 READ          1 HELD    test.dbxml                page         13
    80000045 WRITE         1 WAIT    test.dbxml                page         13
    80000054 READ          1 WAIT    test.dbxml                page         13
    80000055 READ          1 WAIT    test.dbxml                page         13
    
          b5 READ          1 HELD    test.dbxml                handle       12
    
          a1 READ          1 HELD    test.dbxml                handle        2
    
          b5 READ          6 HELD    test.dbxml                handle        0
          c0 READ          2 HELD    test.dbxml                handle        0
          f1 READ          1 HELD    test.dbxml                handle        0
    
          a9 READ          1 HELD    test.dbxml                handle        6
    
          a5 READ          1 HELD    test.dbxml                handle        4
    
          f1 READ          1 HELD    test.dbxml                handle       18
    
          c0 READ          1 HELD    test.dbxml                handle       16

    The problem here is that you have multiple active transactions on a single thread. The reason that the transaction wire 80000051 2648 is not moving forward and renounce its lock is because the thread is blocked with transaction 80000057 2648 thread waiting on page 11, that 80000045 holds. When a thread has several transactions in Berkeley DB XML built detection of blockages can not say he is in a bind, because it assumes that transactions are on different threads and therefore increase.

    Lauren Foutz

  • Packaging is unable to create the output folder: deliverables/Standard/5.0.0

    Hi all

    Please help me. I am beginner. And I join 3 different pictures based on the same projects. I mean that eclipse was working thick. But, I close my pc, then after some time, I restart my eclipse. He can't run my blackberry application. It gives this error

    "Packaging could not create the output folder: deliverables/Standard/5.0.0.

    Why?

    Thanks in advance.

    Finally, I solved it. I don't know, but I think this way is the solution.

    (1) eclipse, and javaw.exe or java.exe, I close.

    (2) even once open eclipse,

    (3) clean all projects.

    (4) I have apply this action.

    (5) it works.

  • Unable to understand the syntax index

    Hello
    I can't understand the meaning of the following syntax to index... and how it works
    CREATE UNIQUE INDEX X_GRS_DMSC_U02 ON Ta_GR_MSC
    (DECODE("GR_TP",'U',NULL,"ON_ID"), DECODE("GR_TP",'U',NULL,"GRS_MSC"))
    Why the syntax above is used... And how does it work?

    Could you please update me as soon as possible...

    It looks like a basic index function. It is designed to ensure that there is a unique combination of the columns 'ON_ID' AND 'GRS_MSC', if the value in the "GR_TP" column is not equal to "U".

    I will try to create the 'TA_GRS_MSC' table and apply the index to illustrate the above as follows:

    SQL> create table ta_gr_msc (gr_tp varchar2(10),on_id varchar2(10), grs_msc varchar2(10));
    
    Table created.
    
    SQL> CREATE UNIQUE INDEX X_GRS_DMSC_U02 ON Ta_GR_MSC
      2  (DECODE("GR_TP",'U',NULL,"ON_ID"), DECODE("GR_TP",'U',NULL,"GRS_MSC"));
    
    Index created.
    
    SQL> insert into ta_gr_msc values ('U','1', '1');
    
    1 row created.
    
    SQL> insert into ta_gr_msc values ('U','1', '1');
    
    1 row created.
    
    SQL> insert into ta_gr_msc values ('X','1','1');
    
    1 row created.
    
    SQL> INSERT INTO TA_GR_MSC VALUES ('X','1','1');
    INSERT INTO TA_GR_MSC VALUES ('X','1','1')
    *
    ERROR at line 1:
    ORA-00001: unique constraint (USER01.X_GRS_DMSC_U02) violated 
    
    SQL> INSERT INTO TA_GR_MSC VALUES ('X','1','2');
    
    1 row created.
    

    As noted above, when the value in the column 'GR_TP' is equivalent to 'U', you can insert any combination of the values in double columns 'ON_ID' and 'GRS_MSC '. However, if the value in the "GR_TP" column is not equal to 'U', you get a unique constraint exception when you try to insert values in double columns 'ON_ID' and 'GRS_MSC '.

    I hope this helps.

  • Unable to understand the Plan to explain

    Hi gurus

    I'm trying to understand some basics of explain plan and get a hard time, I was reading the book tuning performance and incapable of understanding explain plan for the following query:

    Example query

    EXPLAIN PLAN FOR

    SELECT *.

    WCP

    WHERE THERE IS NOT (SELECT 0

    OF THE Department

    WHERE dept.dname = 'SALES' AND dept.deptno = emp.deptno)

    AND NOT EXISTS (SELECT 0

    Bonus OF

    WHERE bonus.ename = emp.ename);

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

    Select * from table (dbms_xplan.display);

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

    Output

    Hash value of plan: 734347697

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

    | ID | Operation | Name | Lines | Bytes | Cost (% CPU). Time |

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

    |   0 | SELECT STATEMENT |       |     5.   290.     7 (15) | 00:00:01 |

    |*  1 |  HASH ANTI JOIN |       |     5.   290.     7 (15) | 00:00:01 |

    |*  2 |   HASH ANTI JOIN |       |     5.   255.     5 (20) | 00:00:01 |

    |   3.    TABLE ACCESS FULL | EMP |    14.   532.     2 (0) | 00:00:01 |

    |*  4 |    TABLE ACCESS FULL | DEPT |     1.    13.     2 (0) | 00:00:01 |

    |   5.   TABLE ACCESS FULL | BONUS |     1.     7.     2 (0) | 00:00:01 |

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

    Information of predicates (identified by the operation identity card):

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

    1 - access("BONUS".") ENAME "=" EMP ". ("' ENAME ')

    2 - access("DEPT".") DEPTNO "=" EMP ". ("' DEPTNO ')

    4 - filter("DEPT".") DNAME "= 'SALES')

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

    Grateful if someone help out me. Thank you

    In addition, I really appreciate if someone proposes a simple tutorial to explain plan. Thanks again

    Concerning

    Shu

    Hi Shuumail,

    Here is the tutorial http://allthingsoracle.com/execution-plans-part-1-finding-plans/

  • Unable to understand the original clause


    Hi gurus

    I tried to understand what follows where the clause, but still facing problem to understand that and I really appreciate if someone explain it to me step by step and in detail. Thanks in advance:

    Where Clause

    NOT the CASE (min_age = min_min_age

    OR (min_age - 1, rule_id, status_code)

    IN (SELECT max_age, rule_id, status_code

    OF got_analytics

    )

    )

    Concerning

    Shu

    not ( -- simple
      min_age = min_min_age  -- simple
      or -- simple
      (min_age-1, rule_id, status_cod) in (select max_age,rule_id,status_cod from analytics) -- complex
    )
    

    Description for COMPLEX line

    The triplicate (min_age-1, rule_id, status_code)

    must match one of the Geometry produced by:

      (select
          max_age,rule_id,status_code -- your triplicate
        from analytics)
    

    You'll have to check that my logic, but I think that the 'complex' line can be re-written as a clause EXISTS

    EXISTS ( select null
             from analytics A
             where A.max_age = X.min_age - 1
               and A.rule_id = X.rule_id
               and A.status_code = X.status_code )
    

    MK

  • Unable to create the output using RoboHelp 8 of Adobe AIR

    Hello!

    I can't create output AIR Adobe RoboHelp 8. When I compile the project, I receive the following error message. Can someone check and let me know what it takes to create the AIR outlet with success?

    AIR Error.jpg

    I'm not familiar with "RoboHelp Packager for AIR RoboHelp" - I always created AIR help using the same presentation Source RH8. Maybe you should try to create a new presentation of SSL in the AIR (do not copy an existing one - create a new one from scratch to eliminate corruption).

  • Cannot install some programs-can't create the output file in the AppData\Local\Temp user account

    When I try to install a program (.exe) it works winzip Self-extractor, which attempts to create a file in the directory Users\ (myusername) \AppData\Local\Temp\setup.exe

    I just installed windows 7 and had to happen one or two times before that, I think, but I re-uploaded the file and I tried again and it ended up working. Now when I try to install this program (Hava), he says that it is unable to create the output file.

    When I try to go to this place it doesn't even have the AppData folder I have to first create this path or is it a problem with the program? Any help is appreciated. Touch Pro

    Well, that's possible. As WinZip SFX has a small option to overwrite without confirmation box.

    You can move forward with delete the file setup.exe from the Temp folder in the sound is no longer used by any program (where its in the Temp folder)

    You can also try now is, since its opening a WinZIP Self Extracting Archive, you can try manually, and manually specify where to extract the files before you click ok.

    Once you do this, just run setup.exe from the specified location.

  • How to get the output of a compiled solution?

    I'm writing an automation, but I'm unable to get the output of devenv, but receive nothing.  I have reduced my example considerably.

    I have a lot of files that looks like this

    "B:\Microsoft Visual Studio 9.0\Common7\IDE\devenv" "My_Types\My_Types.sln" /build >... \Output\Build.txt

    And it produces Build.txt

    Microsoft (R) Visual Studio Version 9.0.21022.8.
    Copyright (C) Microsoft Corp. All rights reserved.
    -Build started: project: Rotair_Types, Configuration: Debug Any CPU.
    Rotair_Types-> C:\IntuitiveDev\Bin\Rotair_Types.dll
    = Build: 1 succeeded or up-to-date, 0 failed, 0 was ignored.

    So now, I try to do it from code:

    Compile = new Process()
    With Compile.StartInfo
    . UseShellExecute = False
    . RedirectStandardOutput = True
    . FileName = "b:\Microsoft Visual Studio 9.0\Common7\IDE\devenv.exe"
    . Arguments = "/build" «C:\Code\Projects\My_Types\My_Types.sln»»»
    Ends with
    bSuccess = Compile.Start)
    strOutput = Compile.StandardOutput.ReadToEnd)
    Compile.WaitForExit)

    But strOutput is an empty string.  This is confirmed by the issuance of the following at the command prompt:

    /build "C:\Code\Projects\My_Types\My_Types.sln" "b:\Microsoft Visual Studio 9.0\Common7\IDE\devenv.exe"

    But it does not compile because I can remove the dll, and it returns.  I would get the result to determine if the compilation is successful.  Any suggestions?

    JM

    Hello

    Your question is addressed in these forums of consumers.

    The MSDN Developer forums should be able to answer your question.

    Issues related to the overall development of the Windows Desktop:

    https://social.msdn.Microsoft.com/forums/en-us/home?category=windowsdesktopdev

    Concerning

  • You cannot change the output type and format in the page requests system - PeopleSoft process

    Hi all

    When I navigate to the PeopleSoft process scheduler page and try to run a process I see the output type and format web and txt only. I am unable to change the output type and format. All formats and types of output are marked as active in the system settings page. I'm super user of the system as well. I have as well the role of Admin ProcessScheduler. I would like to know if someone has already faced similar problems.

    Thank you

    Vimal

    Run the following SQL query to your OPRID:

    SELECT A.CLASSID, PSPRCSPRFL A.OVRDOUTDEST, PSOPRDEFN B WHERE B.PRCSPRFLCLS = A.CLASSID AND B.OPRID = '';

    Make sure that OVRDOUTDEST = 1.

    If it's 0 then you have to fix your configuration profile of the process as follows:

    -Main menu > PeopleTools > Security > list permission > CLASSID research (white list) back to top query

    -Go to the process tab

    -Click on "process the profile permissions

    S ' ensure that 'substitute an output Destination' is checked. Click OK, and then save.

    This should solve your problem.

Maybe you are looking for

  • Canon Pixma iP1500 printer and it connect not gives an error code 20

    I have a canon Pixma iP1500 printer and it does not connect gives an error code 20?

  • Restart the installation wizard

    My environment is changing. What is the best way to re - run the Setup Wizard?

  • Persistent policy object

    Is it advisable to keep my own app to the data store objects or use a hash table? For example, I have a PERSON object with name & I still there. In the future, the object may change. Will this be a problem retrieving in the store? If I persisit a has

  • HP pavilion 038sa-15 driwers for win7

    Hello eweryone can someone help me find drivers for my laptop but for win7, I reale hate win8 and does not usually use it. So installed win 7 but no network device is found, tried to install other drivers drank without success. I know that this lapto

  • Affinity DRS rules?

    Hello! I use Affinity DRS rules at present with DRS groups etc. My question is if it is possible for a little more advanced with the DRS group management? Is it possible to group together (for example tag)-based VM rather than manually? So ideally, I