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

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?

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.

Most likely because you're not including the header file for malloc and, while the compiler would normally warn you of this, the fact that you're explicitly casting the return value means you're telling it you know what you're doing.

That means the compiler expects an int to be returned from malloc which it then casts to a pointer. If they're different sizes, that's going to cause you grief.

This is why you never cast the malloc return in C. The void* that it returns will be implicitly converted to a pointer of the correct type (unless you haven't included the header in which case it probably would have warned you of the potentially unsafe int-to-pointer conversion).