Size of array when index is zero

int main()
                 {
                      int arr[0];
                      arr[0]=1;
                        printf("size=%d val=%d\n",sizeof(arr),arr[0]);
                        return 0;
                   }

output is 0,1. why its size is zero if there is no memory allocated then how it can store a value.

why its size is zero if there is no memory allocated then
how it can store a value.

mohansaini

"int arr[0]; " creates a pointer to an integer, that is, a location in
memory where an integer can be stored.

The complier It does not actually have to allocate any storage.
The statement is suspicious, but not illegal.

"arr[0]=1;" places '1' at that location.
The location physically exists, but very likely is in use for
something else.