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sscanf causing core dump

Posted on 2004-10-19
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Last Modified: 2010-04-15
I have come across a bit of code that uses sscanf.  The function containing it is as follwos,

void f_extract_n(char *string, SINT *value, short *ind, int len )
{
    char    tmp[30];

    memset(&tmp, 0, 30 );
    strncpy( tmp, string, len );
printf("calling sscanf\n");
    if ( (sscanf(tmp, "%d", (unsigned int) value )) == EOF )
        *ind   = -1;
printf("sscanf ok\n");

    return;
}

When run under AIX 4.3(32bit) it works fine (where value in tmp is "001108") but when run under AIX 5.2(64 bit) the sscanf is core dumping (same value of tmp).  If I remove the (unsigned int) cast it works ok.  Can anyone tell me why the (unsigned int) cast is causing the core dump?
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Question by:DLyall
8 Comments
 
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by:Kent Olsen
ID: 12346623

sscanf needs a target location to place the converted value.  In this case you're recasting an address into an integer!  C is dutifully placing the integer on the stack prior to calling sscanf() and using this value as the address to store the conversion.

I suspect that the recast to an int (32-bits?) is dropping the upper 32 bits of the 64-bit address in the *value parameter.


Kent
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by:stefan73
ID: 12346679
Hi DLyall,
Replace (unsigned int) by (unsigned long).


Cheers!

Stefan
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Accepted Solution

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Kent Olsen earned 2000 total points
ID: 12346795

Hi Stephan,

That is still an implementation specific solution because the pointer is being recast to an integer.  And recasting it to be a pointer to an integer (or pointer to a long) is no better because the function header explicitly declares the value to be a pointer to a short.  sscanf() could overwrite something that it shouldn't if the variable is improperly recast.

The best solution is to simply throw away the recasting and let the C compiler (and libraries) work.


Kent
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LVL 46

Expert Comment

by:Kent Olsen
ID: 12346809

I suspect that the code should recast the return value of sscanf(), not the pointer being passed.

Kent
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Expert Comment

by:AlexFM
ID: 12347136
Out of topic: I wonder how this is working -

char    tmp[30];
memset(&tmp, 0, 30 );         // should be tmp and not &tmp
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Expert Comment

by:van_dy
ID: 12347332
although memset(&tmp, 0, 30) is used wrongly, it can still work, only zeroing some other memory
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Expert Comment

by:Kent Olsen
ID: 12347378

I believe that the compiler will simply disregard the unnecessarey '&'.  Depending on the compiler options, it should produce a "superfluous '&'" warning.

Kent

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by:DLyall
ID: 12347927
Cheers Kent

Removing the cast works on both the 32 and 64 bit environments.  Also I output the values of &tmp and tmp and they are the same so it looks like the & is ignored at compile time.

Thanks for your Help

Don
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