Why does this code segfault on 64-bit architecture but work fine on 32-bit?

c pointers segmentation-fault
user7 picture user7 · Sep 25, 2011 · Viewed 12.9k times · Source

I came across the following C puzzle:

Q: Why does the following program segfault on IA-64, but work fine on IA-32?

  int main()
  {
      int* p;
      p = (int*)malloc(sizeof(int));
      *p = 10;
      return 0;
  }

I know that the size of int on a 64 bit machine may not be the same as the size of a pointer (int could be 32 bits and pointer could be 64 bits). But I am not sure how this relates to the above program. Any ideas?

Answer

Flexo picture Flexo · Sep 25, 2011

The cast to int* masks the fact that without the proper #include the return type of malloc is assumed to be int. IA-64 happens to have sizeof(int) < sizeof(int*) which makes this problem obvious.

(Note also that because of the undefined behaviour it could still fail even on a platform where sizeof(int)==sizeof(int*) holds true, for example if the calling convention used different registers for returning pointers than integers)

The comp.lang.c FAQ has an entry discussing why casting the return from malloc is never needed and potentially bad.

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