Question

#include <stdio.h>

enum {AA, BB, CC} s;

int main()
{
s = 4;
printf("%d\n",s);
return 0;
}

The compiler doesn’t give any warning and prints 4. What is happening behind the scene? Is s treated as an int type?

Answer 1

The specific type of an enumeration is implementation specific, but it is often an int. So yes, in this case s is probably an int. From the C spec:

Each enumerated type shall be compatible with char, a signed integer type, or an unsigned integer type. The choice of type is implementation-defined, but shall be capable of representing the values of all the members of the enumeration. The enumerated type is incomplete until after the } that terminates the list of enumerator declarations.

So in your case, 4 will certainly work, since it fits in a char and in any signed or unsigned integer type on any machine I’ve ever heard of.