问题是如何将wstring转换为字符串?

我还有一个例子:

#include <string>
#include <iostream>

int main()
{
    std::wstring ws = L"Hello";
    std::string s( ws.begin(), ws.end() );

  //std::cout <<"std::string =     "<<s<<std::endl;
    std::wcout<<"std::wstring =    "<<ws<<std::endl;
    std::cout <<"std::string =     "<<s<<std::endl;
}

带注释的输出为:

std::string =     Hello
std::wstring =    Hello
std::string =     Hello

但是without只是:

std::wstring =    Hello

这个例子中有什么问题吗?我可以像上面那样进行转换吗?

EDIT

新例子(考虑到一些答案)是

#include <string>
#include <iostream>
#include <sstream>
#include <locale>

int main()
{
    setlocale(LC_CTYPE, "");

    const std::wstring ws = L"Hello";
    const std::string s( ws.begin(), ws.end() );

    std::cout<<"std::string =     "<<s<<std::endl;
    std::wcout<<"std::wstring =    "<<ws<<std::endl;

    std::stringstream ss;
    ss << ws.c_str();
    std::cout<<"std::stringstream =     "<<ss.str()<<std::endl;
}

输出结果为:

std::string =     Hello
std::wstring =    Hello
std::stringstream =     0x860283c

因此,不能使用stringstream将wstring转换为string。


当前回答

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。

其他回答

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; }

而不是包括locale和所有那些花哨的东西,如果你知道为FACT你的字符串是可转换的,只需这样做:

#include <iostream>
#include <string>

using namespace std;

int main()
{
  wstring w(L"bla");
  string result;
  for(char x : w)
    result += x;

  cout << result << '\n';
}

这里有一个活生生的例子

我花了很多悲伤的日子,试图为c++ 17找到一种方法来做到这一点,它已经弃用了code_cvt facet,这是我通过组合来自几个不同来源的代码所能想到的最好的方法:

setlocale( LC_ALL, "en_US.UTF-8" ); //Invoked in main()

std::string wideToMultiByte( std::wstring const & wideString )
{
     std::string ret;
     std::string buff( MB_CUR_MAX, '\0' );

     for ( wchar_t const & wc : wideString )
     {
         int mbCharLen = std::wctomb( &buff[ 0 ], wc );

         if ( mbCharLen < 1 ) { break; }

         for ( int i = 0; i < mbCharLen; ++i ) 
         { 
             ret += buff[ i ]; 
         }
     }

     return ret;
 }

 std::wstring multiByteToWide( std::string const & multiByteString )
 {
     std::wstring ws( multiByteString.size(), L' ' );
     ws.resize( 
         std::mbstowcs( &ws[ 0 ], 
             multiByteString.c_str(), 
             multiByteString.size() ) );

     return ws;
 }

我在Windows 10上测试了这段代码,至少就我的目的而言,它似乎运行良好。如果这没有考虑到你可能需要处理的一些疯狂的边缘情况,请不要对我进行私刑,我相信有更多经验的人可以改进这一点!: -)

此外,在该表扬的地方表扬:

适用于wideToMultiByte()

复制multiByteToWide

正如Cubbi在一条评论中指出的那样,std::wstring_convert (c++ 11)提供了一个简洁的解决方案(你需要#include <locale>和<codecvt>):

std::wstring string_to_convert;

//setup converter
using convert_type = std::codecvt_utf8<wchar_t>;
std::wstring_convert<convert_type, wchar_t> converter;

//use converter (.to_bytes: wstr->str, .from_bytes: str->wstr)
std::string converted_str = converter.to_bytes( string_to_convert );

在遇到这个问题之前,我正在使用wcstombs和繁琐的内存分配/释放的组合。

http://en.cppreference.com/w/cpp/locale/wstring_convert

更新(2013.11.28)

有一句话可以这样说(谢谢你的评论):

std::wstring str = std::wstring_convert<std::codecvt_utf8<wchar_t>>().from_bytes("some string");

包装器函数可以这样表述:(感谢ArmanSchwarz的评论)

std::wstring s2ws(const std::string& str)
{
    using convert_typeX = std::codecvt_utf8<wchar_t>;
    std::wstring_convert<convert_typeX, wchar_t> converterX;

    return converterX.from_bytes(str);
}

std::string ws2s(const std::wstring& wstr)
{
    using convert_typeX = std::codecvt_utf8<wchar_t>;
    std::wstring_convert<convert_typeX, wchar_t> converterX;

    return converterX.to_bytes(wstr);
}

注意:对于string/wstring是否应该作为引用或文字传递给函数存在一些争议(由于c++ 11和编译器更新)。我将把决定留给执行的人,但这是值得了解的。

注意:我在上面的代码中使用std::codecvt_utf8,但如果你不使用UTF-8,你需要将其更改为你正在使用的适当编码:

http://en.cppreference.com/w/cpp/header/codecvt

如果你正在处理文件路径(当我发现需要wstring-to-string时,我经常这样做),你可以使用filesystem::path(自c++ 17以来):

#include <filesystem>

const std::wstring wPath = GetPath(); // some function that returns wstring
const std::string path = std::filesystem::path(wPath).string();