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cmemcpy

Why does compiler not give a warning in this case

This piece of code :

       char buff[255];
       memcpy(buff,0,255);

The compiler doesn't give any warning during compilation but the process fails with a segmentation fault But when I compile the following code 

       char buff[255];
       memcpy(buff,2,255);

The compiler gives the following warning

warning: passing argument 2 of 'memcpy' makes pointer from integer without a cast [enabled by default]

Why does compiler not give an warning for the constant 0 I am using GCC version 4.7.2

Also is there some compiler flag that will give warnings for such code

Solution

  • 0 is implicitly converted to the null pointer when used in a pointer context. 2 isn't implicitly converted to a pointer type, hence the warning.

    You can read more at the comp.lang.c FAQ, section 5, specifically Question 5.2.

    In a quick test here, neither GCC nor Clang warned about the 0 case (without extra flags), but the clang static analyzer does:

    example.c:6:4: warning: Null pointer argument in call to memory copy function
   memcpy(buff,0,255);
   ^~~~~~~~~~~~~~~~~~
/usr/include/secure/_string.h:55:6: note: expanded from macro 'memcpy'
   ? __builtin___memcpy_chk (dest, src, len, __darwin_obsz0 (dest))     \
     ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1 warning generated.

    As has been mentioned elsewhere in comments here, GCC will warn if you pass -Wnonnull (also included with -Wall):

    $ gcc -Wnonnull example.c -o example
example.c: In function ‘main’:
example.c:6: warning: null argument where non-null required (argument 2)