Plus sign "+" before wchar_t variable

Question

I have found on the cpp reference website (link) the following code which I do not understend completely:

void try_widen(const std::ctype<wchar_t>& f, char c)
{
    wchar_t w = f.widen(c);
    std::cout << "The single-byte character " << +(unsigned char)c
              << " widens to " << +w << '\n';
}

What I do not understand is what does plus sign "+" makes before w variable (+w) and before (unsigned char)c (+(unsigned char)c). Can somebody explain me this?

I believe it forces a conversion to an integer instead of a character. — Retired Ninja, May 23 '22 at 22:42

score 2 · Accepted Answer · answered May 23 '22 at 22:47

2

Like all arithmetic operations, unary plus triggers integer promotion of its operand. This results in printing a numerical value of a character. Without it, a different overload of operator << could be chosen. << (unsigned char)c would print a character rather than a number. I believe std::cout << w would not compile.

answered May 23 '22 at 22:47

Eugene

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Yes you are right `std::cout << w` does not compile but `std::wcout << w` does compile. – joe_specimen May 24 '22 at 09:10

Plus sign "+" before wchar_t variable

1 Answers1