An amount of maximum code lines exist?

Hello

Some time ago, I posted this thread: http://forums.Adobe.com/thread/983062 and nobody has been able to help me yet.

I started to do a lot of testing, which notably to make a copy of the project data and commenting on some of its parts. After some time trying to find a model to the error, I noticed, that the construction of the ipa process was successful not according to what I comment out but on how much.

My main function has 2416 lines of code. Earlier, I tried commenting on about 200 lines and it came out, it didn't matter if I did at the beginning of a certain function, or the end of it.

The ipa was built successfully without the 200 lines at the beginning or 200 lines at the end. But when I leave the complete code, I get the error out-of-memory such as mentioned in the thread the link that I posted above.

The other idea is that maybe there is a limit of memory to save data in the tables. I use two-dimensional arrays and function in that I use nearly consists of 2000 lines with code, storying data in them. There are a 18 paintings together with 3 medium subtables that store about 20 entries each, resulting in cells of table filled about 1080. 200 lines, that I made comments are part of 1/8 of the whole function.

So now the three questions are:

1. is there a limit to the lines of code?

2. is there a limit to the tables/table entries?

3. is there a way to increase the memory for the ipa, construction process?

Thank you very much

GDdept

Seems ok I found a limit. The limit is of about how strings can be stored in the tables within a function. I separated just the function in 6 functions and the building has operated.

Tags: Adobe AIR

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    SELECT NULL l1, 'T' l2, 'Y' l3, NULL l4, 'J' l5 FROM dual UNION ALL
    
    SELECT 'T' l1, 'T' l2, 'R' l3, 'E' l4, NULL l5 FROM dual UNION ALL
    SELECT 'W' l1, 'T' l2, NULL l3, 'G' l4, NULL l5 FROM dual 
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    TABLE2 AS
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    SELECT 2 cid,'K' l1, 'L' l2, NULL l3, NULL l4, 'I' l5 FROM dual UNION ALL
    SELECT 3 cid,NULL l1, 'T' l2, 'Y' l3, NULL l4, 'J' l5 FROM dual 
    
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    what I want to do is take each row in table1 and whether they exists in table2.
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    CID    l1 l2  l3  l4  l5
    ====  === ==  === === ===
    1      T  T          
    2      K  L            I
    3         T    Y       J
          
           T  T    R   E
           W  T        G
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    If a row from table1 exist in table2 based on all the columns of l, then I view with decided to table2.
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    Select columns
    FROM table1, table2 b
    where a.l1 = b.l1 (+)
    and a.l2 = bl2 (+)
    and...
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    SELECT 'T' l1, 'T' l2, NULL l3, NULL l4, NULL l5 FROM dual UNION ALL
    SELECT 'K' l1, 'L' l2, NULL l3, NULL l4, 'I' l5 FROM dual UNION ALL
    SELECT NULL l1, 'T' l2, 'Y' l3, NULL l4, 'J' l5 FROM dual UNION ALL
    
    SELECT 'T' l1, 'T' l2, 'R' l3, 'E' l4, NULL l5 FROM dual UNION ALL
    SELECT 'W' l1, 'T' l2, NULL l3, 'G' l4, NULL l5 FROM dual
    ),
    TABLE2 AS
    (
    SELECT 1 cid,'T' l1, 'T' l2, NULL l3, NULL l4, NULL l5 FROM dual UNION ALL
    SELECT 2 cid,'K' l1, 'L' l2, NULL l3, NULL l4, 'I' l5 FROM dual UNION ALL
    SELECT 3 cid,NULL l1, 'T' l2, 'Y' l3, NULL l4, 'J' l5 FROM dual
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    select table2.cid, table1.l1, table1.l2, table1.l3, table1.l4, table1.l5 from
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    nvl(table1.l1,0) = nvl(table2.l1,0) and
    nvl(table1.l2,0) = nvl(table2.l2,0) and
    nvl(table1.l3,0) = nvl(table2.l3,0) and
    nvl(table1.l4,0) = nvl(table2.l4,0) and
    nvl(table1.l5,0) = nvl(table2.l5,0)
    union all
    select null, table1.l1, table1.l2, table1.l3, table1.l4, table1.l5 from
    table1 where not exists (select 1 from table2 where nvl(table1.l1,0) = nvl(table2.l1,0) and
    nvl(table1.l2,0) = nvl(table2.l2,0) and
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    "1"     "T"     "T"     ""     ""     ""
    "2"     "K"     "L"     ""     ""     "I"
    "3"     ""     "T"     "Y"     ""     "J"
    ""     "W"     "T"     ""     "G"     ""
    ""     "T"     "T"     "R"     "E"     ""
    

    Published by: APRIL on February 28, 2011 19:51

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