为什么w / cout不支持字符串U / u前缀？

Question

I have the following code inspired by one of the SO questions:

int main()
{
    wcout << "\n\n 1- Test normal string with wcout \n";
    wcout << u8"\n 2- Test (u8) string with wcout  \n";
    wcout << u"\n 3- Test (u) string with wcout \n";
    wcout << U"\n 4- Test (U) string with wcout \n";
    wcout << L"\n 5- Test (L) string with wcout \n\n"; 
}

which results in:

cout has similar output.

I understand from the answer on this post that w/cout does not support the string U/u prefixes, or exactly as mentioned:

Meanwhile std::cout's class has no special << overload for const char16_t* , const char32_t* and const wchar_t* , hence it will match << 's overload for printing pointers.

My questions are:

1- Why it does not support these types? Isn't it a strange for a language not to support its own types? 2- Are there any known plans to add such support in the seen future?

I know that iostream is a library, but, practically, it is still part of the language.

Answer 1

The usual answer to standard library omissions applies: nobody proposed to add overloads for char16_t const* or char32_t . The proposal to add the character types ( N2249 ) did not contain the relevant library parts. In particular, it didn't add overloads for

template <typename charT, typename Traits>
std::basic_ostream<charT, Traits>& std::operator<< (std::basic_ostream<charT, Traits>&, char16_t const*);
template <typename charT, typename Traits>
std::basic_ostream<charT, Traits>& std::operator<< (std::basic_ostream<charT, Traits>&, char32_t const*);

As there is an overload of these operator taking void const* the address of the pointer is printed. As the new character types are kind of pointless (at least, from my point of view: just use UTF8) I doubt there is anybody interested in creating a proposal.