Page 188 - C-Language

However, if the type generic macro is implemented like this:

#define is_const_int(x) _Generic((x), \
const int: "a const int", \
int: "a non-const int", \
default: "of other type")

The output is:

i is a non-const int
j is a non-const int
k is of other type

This is because all type qualifiers are dropped for the evaluation of the controlling expression of a
_Generic primary expression.

Type-generic printing macro

#include <stdio.h>

void print_int(int x) { printf("int: %d\n", x); }
void print_dbl(double x) { printf("double: %g\n", x); }
void print_default() { puts("unknown argument"); }

#define print(X) _Generic((X), \
int: print_int, \
double: print_dbl, \
default: print_default)(X)

int main(void) {
print(42);
print(3.14);
print("hello, world");
}

Output:

int: 42
double: 3.14
unknown argument

Note that if the type is neither int nor double, a warning would be generated. To eliminate the
warning, you can add that type to the print(X) macro.

Generic selection based on multiple arguments

If a selection on multiple arguments for a type generic expression is wanted, and all types in
question are arithmetic types, an easy way to avoid nested _Generic expressions is to use addition
of the parameters in the controlling expression:

int max_int(int, int);
unsigned max_unsigned(unsigned, unsigned);

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