Nsryjdtyk

#include <stdio.h>

#include <stdbool.h>

typedef enum{

    False,

    True

}Bool;

int main(int argc, char const *argv)

{

    int a = 1, d = 1;

    Bool b = False, c = True;

    Bool ans;

    if (a == d)

    {

        ans = c;



    }

    else

    {

        ans = b;

    }

    printf("The answer is: %i n", ans);

    return 0;

}

I think it is only returning the execution result 0 or 1.
I want it to return the values from the enum which I created.

If you want it to print the string true or false you'll have to add something like this:

printf("The answer is: %sn", ans ? "true" : "false");

From the C Standard, § 6.7.2.2:2-3:

The identifiers in an enumerator list are declared as constants that have type int and
may appear wherever such are permitted.) An enumerator with = defines its
enumeration constant as the value of the constant expression. If the first enumerator has
no =, the value ... is 0. Each subsequent enumerator with no =
defines its ... value ... by
adding 1 to the value of the previous enumeration constant.

Emphasis mine.

In C an enum is no more than an int type but with special symbols for a subset of those ints. Also, unless you tell it otherwise, the implicit int value for the first enum member is 0, the second is 1, and so on.

%i can be used to print an int, which accounts for the output.

It is not possible to display the actual enum token in portable C, although there are idiomatic approaches that make use of the preprocessor.