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如何在 C++ 中 cout wstring 或 wchar_t

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问题：

你有一个 std::wstring 想使用 cout 打印

test-wstring.cpp
wstring w = L"Test: äöü";
cout << w << endl;

但你看到类似这样结尾的长错误消息：

compile-error.txt
/usr/include/c++/7/ostream:682:5: note:   template argument deduction/substitution failed:
/usr/include/c++/7/ostream: In substitution of 'template<class _Ostream, class _Tp> typename std::enable_if<std::__and_<std::__not_<std::is_lvalue_reference<_Tp> >, std::__is_convertible_to_basic_ostream<_Ostream>, std::__is_insertable<typename std::__is_convertible_to_basic_ostream<_Tp>::__ostream_type, const _Tp&, void> >::value, typename std::__is_convertible_to_basic_ostream<_Tp>::__ostream_type>::type std::operator<<(_Ostream&&, const _Tp&) [with _Ostream = std::basic_ostream<char>&; _Tp = std::__cxx11::basic_string<wchar_t>]':
test.cpp:9:13:   required from here
/usr/include/c++/7/ostream:682:5: error: no type named 'type' in 'struct std::enable_if<false, std::basic_ostream<char>&>'

解决方案

你需要使用 std::wcout 而不是 std::cout：

print-wstring.cpp
wstring w = L"Test: äöü";
wcout << w << endl;
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