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

在我的情况下,我必须使用多字节字符(MBCS),我想使用std::string和std::wstring。不能使用c++11。所以我使用mbstowcs和wcstombs。

我与使用new, delete[]做相同的函数,但它比这更慢。

这可以帮助如何:在各种字符串类型之间转换

EDIT

然而,在转换为wstring和源字符串是没有字母和多字节字符串的情况下,它是不工作的。 所以我把wcstombs改成WideCharToMultiByte。

#include <string>

std::wstring get_wstr_from_sz(const char* psz)
{
    //I think it's enough to my case
    wchar_t buf[0x400];
    wchar_t *pbuf = buf;
    size_t len = strlen(psz) + 1;

    if (len >= sizeof(buf) / sizeof(wchar_t))
    {
        pbuf = L"error";
    }
    else
    {
        size_t converted;
        mbstowcs_s(&converted, buf, psz, _TRUNCATE);
    }

    return std::wstring(pbuf);
}

std::string get_string_from_wsz(const wchar_t* pwsz)
{
    char buf[0x400];
    char *pbuf = buf;
    size_t len = wcslen(pwsz)*2 + 1;

    if (len >= sizeof(buf))
    {
        pbuf = "error";
    }
    else
    {
        size_t converted;
        wcstombs_s(&converted, buf, pwsz, _TRUNCATE);
    }

    return std::string(pbuf);
}

编辑使用“MultiByteToWideChar”而不是“wcstombs”

#include <Windows.h>
#include <boost/shared_ptr.hpp>
#include "string_util.h"

std::wstring get_wstring_from_sz(const char* psz)
{
    int res;
    wchar_t buf[0x400];
    wchar_t *pbuf = buf;
    boost::shared_ptr<wchar_t[]> shared_pbuf;

    res = MultiByteToWideChar(CP_ACP, 0, psz, -1, buf, sizeof(buf)/sizeof(wchar_t));

    if (0 == res && GetLastError() == ERROR_INSUFFICIENT_BUFFER)
    {
        res = MultiByteToWideChar(CP_ACP, 0, psz, -1, NULL, 0);

        shared_pbuf = boost::shared_ptr<wchar_t[]>(new wchar_t[res]);

        pbuf = shared_pbuf.get();

        res = MultiByteToWideChar(CP_ACP, 0, psz, -1, pbuf, res);
    }
    else if (0 == res)
    {
        pbuf = L"error";
    }

    return std::wstring(pbuf);
}

std::string get_string_from_wcs(const wchar_t* pcs)
{
    int res;
    char buf[0x400];
    char* pbuf = buf;
    boost::shared_ptr<char[]> shared_pbuf;

    res = WideCharToMultiByte(CP_ACP, 0, pcs, -1, buf, sizeof(buf), NULL, NULL);

    if (0 == res && GetLastError() == ERROR_INSUFFICIENT_BUFFER)
    {
        res = WideCharToMultiByte(CP_ACP, 0, pcs, -1, NULL, 0, NULL, NULL);

        shared_pbuf = boost::shared_ptr<char[]>(new char[res]);

        pbuf = shared_pbuf.get();

        res = WideCharToMultiByte(CP_ACP, 0, pcs, -1, pbuf, res, NULL, NULL);
    }
    else if (0 == res)
    {
        pbuf = "error";
    }

    return std::string(pbuf);
}

其他回答

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。

// 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"

我相信官方的方法仍然是使用codecvt facet(您需要某种语言环境感知的转换),例如

resultCode = use_facet<codecvt<char, wchar_t, ConversionState> >(locale).
  in(stateVar, scratchbuffer, scratchbufferEnd, from, to, toLimit, curPtr);

或者类似的东西,我没有工作代码。但我不确定现在有多少人使用这种机器,有多少人只是要求内存指针,让ICU或其他库处理血腥的细节。

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

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; }