Is there a generic conversion specifier for printf?

Question

I wanna print variable value without specifying its type.

In c, I can do

int main(int argc, char **argv) {
    int i = 1;
    float f = 0.1;
    char s[] = "s";

    printf("%i\n", i);
    printf("%f\n", f);
    printf("%s", s);
    return 0;
}

but I expect:

int main(int argc, char **argv) {
    int i = 1;
    float f = 0.1;
    char s[] = "s";

    printf("%any_type\n", i);
    printf("%any_type\n", f);
    printf("%any_type", s);
    return 0;
}

question: is there %any_type in C?

Answer 1

In C11 you can write a generic function to print any type and keep adding your custom type to that function.

#define print_any(X) _Generic((X), int: print_int, \
                              default: print_unknown, \
                              float: print_float)(X)

int print_int(int i)
{
  return printf("%d\n", i);
}

int print_float(float f)
{
  return printf("%f\n", f);
}

int print_unknown(...)
{
  return printf("ERROR: Unknown type\n");
}

You can also automate the function generation as shown below.

#define GEN_PRINT(TYPE, SPECIFIER_STR) int print_##TYPE(TYPE x) { return printf(SPECIFIER_STR "\n", x);}
GEN_PRINT(int, "%d");
GEN_PRINT(char, "%c");
GEN_PRINT(float, "%f");

Usage will be:

int main(int argc, char **argv) {
    int i = 1;
    float f = 0.1;
    char s[] = "s";

    print_any(i);
    print_any(f);
    print_any(s);
    return 0;
}

Answer 2

No, you have to give the correct format specifier.

In C11 you can automate the computation of the correct format specifier by using a _Generic combined with a macro. However, you have to repeat the name of the variable (once to compute the specifier, and once to give the variable as argument).

For more details read this link.