Question

Assuming that I have this:

enum { A = 0x2E, B = 0x23, C = 0x40 }

it’s possible check if x is defined into enum?

I’m doing it manually: int isdef = (x == A || x == B || x == C); But I want to something more dynamic. GCC-extensions are welcome too.

Answer 1

This is kind of a modified version of your question, but depending on what you’re doing, something like this might work:

enum {A,B,C};
const int E[] = {0x2E,0x23,0x40};
// Or:
// enum { A = 0x2E, B = 0x23, C = 0x40 };
// const int E[] = {A,B,C};

int isEnum(int x)
{
    for(int i=0; i<(sizeof(E)/sizeof(*E)); i++)
    {
        if(E[i] == x){ return 1; }
    }
    return 0;
}

int main(void)
{
    printf("Value of A: 0x%02x\n", E[A]);
    // Or:   
    // printf("Value of A: 0x%02x\n", A);

    printf("isEnum(0x2e): %s\n", isEnum(0x2e) ? "true" : "false");
    printf("isEnum(0x2f): %s\n", isEnum(0x2f) ? "true" : "false");
}

which outputs

Value of A: 0x2e
isEnum(0x2e): true
isEnum(0x2f): false

EDIT: TJD beat me to it, and his suggestion of using a sorted array and doing binary search would decrease your search time from n to log(n).