What did GCC do to the statement of int *a = {1,2,3,4,5}?

Question

#include <stdio.h>
int main(){
  int *a = {1,2,3,4,5};
  printf("a:%x &a:%x\n",a,&a);
  return 0;
}

I compiled this program with GCC. The output of a is 1, and the output of &a is an address. What GCC did to int *a = {1,2,3,4,5}? Did GCC treat it as an array or a pointer that points to an array or something else?

Not because we can't read image of code, but because they are code and should be posted as such. Someone may also want to copy-and-paste it to run it. — user202729, Mar 23 '18 at 01:57
[Array automatically decay into pointer](https://stackoverflow.com/questions/1461432/what-is-array-decaying) in most cases. <--- possible dupe — user202729, Mar 23 '18 at 01:58
comma works as an operator here. that's why you get a=1 as a={1,2,3,4,5} becomes equivalent to “(a = 1),2,3,4,5" . and *a is pointing to an integer. — itsrajon, Mar 23 '18 at 02:05
A pointer to an integer can also be a pointer to the first element of an array. The compiler definitely won't just ignore curly braces. This is not the problem. — eesiraed, Mar 23 '18 at 02:54
After some "what does this code do" like [this one](https://stackoverflow.com/questions/13603286/how-is-int-main-valid-c) it gets quickly boring. In particular this case the compiler basically already explained it. — user202729, Mar 23 '18 at 06:29

pynexj · Answer 1 · 2018-03-23T02:49:15.457

I did not refer to the C standards but you can see how gcc handles this from the compile warning messages:

[STEP 101] # cat foo.c
int main()
{
    int * a = {1, 2};
    return !!a;
}
[STEP 102] # gcc -Wall foo.c
foo.c: In function ‘main’:
foo.c:3:16: warning: initialization makes pointer from integer
without a cast [-Wint-conversion]
     int * a = {1, 2};
                ^
foo.c:3:16: note: (near initialization for ‘a’)
foo.c:3:19: warning: excess elements in scalar initializer
     int * a = {1, 2};
                   ^
foo.c:3:19: note: (near initialization for ‘a’)
[STEP 103] #

UPDATE:

Just took a look at the C99 standard and found this in section 6.7.8 Initialization:

The initializer for a scalar shall be a single expression, optionally enclosed in braces.

score 1 · Answer 2 · answered Mar 23 '18 at 06:21

The code is a constraint violation as an initializer for a non-aggregate can only contain 1 element.

GCC has an "extension" to ignore excess initializers, so it treats the code as int *p = 1;. This is also a constraint violation, because an integer cannot be assigned to a pointer. But gcc has another "extension" to treat such code as int *p = (int *)1;. So you end up with a pointer to address 1.

and the output of &a is an address.

&a is the address in memory of the variable a, this has nothing to do with what value is stored in a.

What did GCC do to the statement of int *a = {1,2,3,4,5}?

2 Answers2