在写这个答案的时候，第一个谷歌搜索“转换字符串wstring”会让你进入这个页面。我的回答展示了如何将字符串转换为wstring，虽然这不是实际的问题，我应该删除这个答案，但这被认为是糟糕的形式。您可能希望跳转到此StackOverflow答案，该答案现在的排名高于此页面。

这是一种将字符串，wstring和混合字符串常量组合到wstring的方法。使用wstringstream类。

#include <sstream>

std::string narrow = "narrow";
std::wstring wide = "wide";

std::wstringstream cls;
cls << " abc " << narrow.c_str() << L" def " << wide.c_str();
std::wstring total= cls.str();

在写这个答案的时候，第一个谷歌搜索“转换字符串wstring”会让你进入这个页面。我的回答展示了如何将字符串转换为wstring，虽然这不是实际的问题，我应该删除这个答案，但这被认为是糟糕的形式。您可能希望跳转到此StackOverflow答案，该答案现在的排名高于此页面。

这是一种将字符串，wstring和混合字符串常量组合到wstring的方法。使用wstringstream类。

#include <sstream>

std::string narrow = "narrow";
std::wstring wide = "wide";

std::wstringstream cls;
cls << " abc " << narrow.c_str() << L" def " << wide.c_str();
std::wstring total= cls.str();

// Embarcadero C++ Builder 

// convertion string to wstring
string str1 = "hello";
String str2 = str1;         // typedef UnicodeString String;   -> str2 contains now u"hello";

// convertion wstring to string
String str2 = u"hello";
string str1 = UTF8string(str2).c_str();   // -> str1 contains now "hello"

这个解决方案的灵感来自dk123的解决方案，但是使用了一个依赖于地区的codecvt facet。结果是区域编码的字符串而不是UTF-8(如果它没有设置为区域设置):

std::string w2s(const std::wstring &var)
{
   static std::locale loc("");
   auto &facet = std::use_facet<std::codecvt<wchar_t, char, std::mbstate_t>>(loc);
   return std::wstring_convert<std::remove_reference<decltype(facet)>::type, wchar_t>(&facet).to_bytes(var);
}

std::wstring s2w(const std::string &var)
{
   static std::locale loc("");
   auto &facet = std::use_facet<std::codecvt<wchar_t, char, std::mbstate_t>>(loc);
   return std::wstring_convert<std::remove_reference<decltype(facet)>::type, wchar_t>(&facet).from_bytes(var);
}

我一直在找，但我找不到。最后，我发现我可以从std::locale使用std::use_facet()函数与正确的typename获得正确的facet。希望这能有所帮助。

如果其他人感兴趣的话:我需要一个可以在任何期望使用string或wstring的地方互换使用的类。下面的类convertible_string，基于dk123的解决方案，可以用string, char const*， wstring或wchar_t const*进行初始化，并且可以被赋值或隐式转换为string或wstring(因此可以传递给接受其中任何一种的函数)。

class convertible_string
{
public:
    // default ctor
    convertible_string()
    {}

    /* conversion ctors */
    convertible_string(std::string const& value) : value_(value)
    {}
    convertible_string(char const* val_array) : value_(val_array)
    {}
    convertible_string(std::wstring const& wvalue) : value_(ws2s(wvalue))
    {}
    convertible_string(wchar_t const* wval_array) : value_(ws2s(std::wstring(wval_array)))
    {}

    /* assignment operators */
    convertible_string& operator=(std::string const& value)
    {
        value_ = value;
        return *this;
    }
    convertible_string& operator=(std::wstring const& wvalue)
    {
        value_ = ws2s(wvalue);
        return *this;
    }

    /* implicit conversion operators */
    operator std::string() const { return value_; }
    operator std::wstring() const { return s2ws(value_); }
private:
    std::string value_;
};

代码有两个问题:

The conversion in const std::string s( ws.begin(), ws.end() ); is not required to correctly map the wide characters to their narrow counterpart. Most likely, each wide character will just be typecast to char. The resolution to this problem is already given in the answer by kem and involves the narrow function of the locale's ctype facet. You are writing output to both std::cout and std::wcout in the same program. Both cout and wcout are associated with the same stream (stdout) and the results of using the same stream both as a byte-oriented stream (as cout does) and a wide-oriented stream (as wcout does) are not defined. The best option is to avoid mixing narrow and wide output to the same (underlying) stream. For stdout/cout/wcout, you can try switching the orientation of stdout when switching between wide and narrow output (or vice versa): #include <iostream> #include <stdio.h> #include <wchar.h> int main() { std::cout << "narrow" << std::endl; fwide(stdout, 1); // switch to wide std::wcout << L"wide" << std::endl; fwide(stdout, -1); // switch to narrow std::cout << "narrow" << std::endl; fwide(stdout, 1); // switch to wide std::wcout << L"wide" << std::endl; }