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();

其他回答

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。

// 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"

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

我使用下面的wstring转换为字符串。

std::string strTo;
char *szTo = new char[someParam.length() + 1];
szTo[someParam.size()] = '\0';
WideCharToMultiByte(CP_ACP, 0, someParam.c_str(), -1, szTo, (int)someParam.length(), NULL, NULL);
strTo = szTo;
delete szTo;

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; }