Wide character output result

Question

Why do I get numbers as output result when using wchar_t?

#include <iostream>
using namespace std;
int main()
{
    wchar_t q;
    q = 'A';
    cout << q;
    q = 'B';
    cout << q;
    q = 'C';
    cout << q;
    return 0;
}

I'm not 100% sure what the rules are, but `wchar_t` can be treated like an integer or something, so that's why you get numbers - because there's no `operator<<` for `wchar_t` with `std::ostream`, so it converts to an integer and uses that overload instead. — chris, Dec 08 '12 at 21:10
Anyway, Im new to C++ so even this is enough for me, for now :) thanks for the info :) — user1888353, Dec 08 '12 at 21:12
Be sure to use `L'A'` and `L"abc"` when using wide characters as well. — chris, Dec 08 '12 at 21:12
You should see http://stackoverflow.com/questions/1613494/why-was-wchar-t-invented — Anil Arya, Dec 08 '12 at 21:12

score 3 · Accepted Answer · answered Dec 08 '12 at 21:26

3

The displayed 'numbers' are the value of the character. That's the reason you get 'numbers'. If you want to display the characters you can use wcout

answered Dec 08 '12 at 21:26

Blazer

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score 0 · Answer 2 · answered Dec 08 '12 at 22:22

std::cout is a basic_iostream<char>. There is no operator<< overload for wchar_t for basic_iostream<char>. When you output wchar_t objects they must be converted to some type for which there is an operator<< overload. The selected conversion is to an integral type other than char and so the integral value is shown instead of the represented character.

Wide character output result

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