Why is the function overload that takes an int preferred over the one taking an unsigned char?

Question

Consider this program:

#include <iostream>

using namespace std;

void f(unsigned char c) {
  cout << c << endl;
}

void f(int c) {
  cout << c << endl;
}

int main() {
  f('a');
}

This prints out 97, suggesting that the f() overload that was selected was the one taking an int. I find this weird; wouldn't intuitively an unsigned char be a better match for a char?

Answer 1

wouldn't intuitively an unsigned char be a better match for a char?

Well, I guess, but not according to the Standard. According to [conv.prom]p1:

A prvalue of an integer type other than bool, char16_­t, char32_­t, or wchar_­t whose integer conversion rank is less than the rank of int can be converted to a prvalue of type int if int can represent all the values of the source type; [...]

Now, the three character types have the same rank, and a signed type has a rank always less than int. This is a combination of [conv.rank]p1.6 and [conv.rank]p1.2:

• The rank of a signed integer type shall be greater than the rank of any signed integer type with a smaller size.

• [...]

• The rank of char shall equal the rank of signed char and unsigned char.

Basically, every character has always a smaller rank than int and they can all be represented in an int, and so the overload with unsigned char is not a better match, because it would involve a conversion from char to unsigned char, instead of a promotion.

If you change your overload to take a char, then there would be an exact match, and so naturally, the "correct" overload (in your eyes) would be chosen.