What is the point of allowing an identifier for enum?

Question

Why doesn't the compiler complain when I try to assign incorrect values to variable a of type enum answer?

#include <stdio.h>

int main()
{
    enum answer {NO, YES};
    enum gender {MALE, FEMALE};

    enum answer a = 5; /* Assign an invalid value. */
    printf("answer: %d\n", a);

    a = MALE; /* Assign a value of wrong type */
    printf("answer: %d\n", a);

    return 0;
}

Here is the output:

$ gcc -std=c99 -pedantic -Wall -Wextra enum.c 
$ ./a.out
answer: 5
answer: 0

If enum doesn't lead to type-checking, then what is the point of having the syntax as:

enum [identifier] {enumerator-list}

I used answer and gender as the identifier for my enum. What is the point of allowing this syntax?

I mean this code could be very well written as

enum {NO, YES};
enum {MALE, FEMALE};

What is the point of allowing this syntax?

enum answer {NO, YES};
enum gender {MALE, FEMALE};

Answer 1

Why doesn't the compiler complain when I try to assign incorrect values to variable a of type enum answer?

Because in C, an enum is practically equivalent to an int. It was standarized like it, and simply too many programs rely on that behavior for it to change.

In C++, they are distinct types and the compiler complains:

$ g++ -Wall -Wextra a.c
a.c: In function 'int main()':
a.c:8:24: error: invalid conversion from 'int' to 'main()::answer' [-fpermissive]
a.c:11:14: error: cannot convert 'main()::gender' to 'main()::answer' in assignment

What is the point of allowing this syntax?

My quick guess would be forward compatibility.

Answer 2

C exposes enumeration values directly as integer while in C++ enum is a real type. Hence in C++ enum results in a type check while in C enum just represents constants of type int. Therefore integer and enum values can be intermixed in all the arithmetic operations.