Question

I have code in C (linux(x86_64)) some like this:

typedef struct 
{
   char k[32];
   int v;
}ABC;

ABC states[6] = {0};

ABC* get_abc()
{
   return &states[5];
}  

while in main():

int main()
{
  ABC *p = get_abc();
     .
     .
     .
  printf("%d\n", p->v); 
} 

I am getting segmentation fault at printf statement while accessing p->v. I tried to debug it from gdb and it says “can not access the memory”. One important thing here is that when I compile this code, gcc throws me a warning on ABC *p = get_abc(); that I am trying to convert pointer from integer. My question here is that I am returning address of structure from get_abc() then why compiler gives me such warning? why compiler considers it as integer? I think I am getting segmentation fault due to this warning as an integer can not hold memory address in x86_64.

Any help would be appreciated.

Answer 1

Define the get_abc prototype before main function. If function prototype is not available before that function call means, compiler will treat that function by default as passing int arguments and returning int. Here get_abc actually returning 8 byte address, but that value has been suppressed to 4 bytes and it is stored in ABC *p variable which leads the crash.

ABC* get_abc(); 
int main()
{
    ABC *p = get_abc();
}

Note : This crash will not occur in 32 bit machine where size of int and size of address is 4 bytes, because suppression will not happen. But that warning will be there.