Check whether a variable is of a certain qualified type
suggest change#include <stdio.h>
#define is_const_int(x) _Generic((&x), \
const int *: "a const int", \
int *: "a non-const int", \
default: "of other type")
int main(void)
{
const int i = 1;
int j = 1;
double k = 1.0;
printf("i is %s\n", is_const_int(i));
printf("j is %s\n", is_const_int(j));
printf("k is %s\n", is_const_int(k));
}
Output:
i is a const int
j is a non-const int
k is of other type
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.
Found a mistake? Have a question or improvement idea?
Let me know.
Table Of Contents