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

来自http://forums.devshed.com/c-programming-42/wstring-to-string-444006.html的旧解决方案

std::wstring wide( L"Wide" ); 
std::string str( wide.begin(), wide.end() );

// Will print no problemo!
std::cout << str << std::endl;

更新(2021年):但是,至少在MSVC的最新版本上,这可能会生成wchar_t到char截断警告。警告可以通过使用std::transform代替转换函数中的显式转换来消除,例如:

std::wstring wide( L"Wide" );

std::string str;
std::transform(wide.begin(), wide.end(), std::back_inserter(str), [] (wchar_t c) {
    return (char)c;
});

或者如果你喜欢预分配而不使用back_inserter:

std::string str(wide.length(), 0);
std::transform(wide.begin(), wide.end(), str.begin(), [] (wchar_t c) {
    return (char)c;
});

参见这里各种编译器的示例。


注意,这里根本没有进行字符集转换。它所做的只是将每个迭代的wchar_t赋值给一个char—截断转换。它使用std::string c'tor:

template< class InputIt >
basic_string( InputIt first, InputIt last,
              const Allocator& alloc = Allocator() );

如评论所述:

值0-127在几乎所有编码中都是相同的,因此要截断 所有小于127的值都会生成相同的文本。输入一个 汉字,你会看到失败。

windows codepage 1252的值128-255 (windows English 默认值)和unicode的值128-255基本相同,因此如果 这就是你所使用的大部分字符的代码页 截断为正确的值。(我完全期待á和õ能正常工作, 我知道我们的代码在工作中依赖于é,我将很快修复)

请注意,Win1252中0x80 - 0x9F范围内的代码点将无法工作。这包括€,,,,…

其他回答

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_;
};

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();

下面是一个基于其他建议的解决方案:

#include <string>
#include <iostream>
#include <clocale>
#include <locale>
#include <vector>

int main() {
  std::setlocale(LC_ALL, "");
  const std::wstring ws = L"ħëłlö";
  const std::locale locale("");
  typedef std::codecvt<wchar_t, char, std::mbstate_t> converter_type;
  const converter_type& converter = std::use_facet<converter_type>(locale);
  std::vector<char> to(ws.length() * converter.max_length());
  std::mbstate_t state;
  const wchar_t* from_next;
  char* to_next;
  const converter_type::result result = converter.out(state, ws.data(), ws.data() + ws.length(), from_next, &to[0], &to[0] + to.size(), to_next);
  if (result == converter_type::ok or result == converter_type::noconv) {
    const std::string s(&to[0], to_next);
    std::cout <<"std::string =     "<<s<<std::endl;
  }
}

这通常适用于Linux,但会在Windows上产生问题。

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; }

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。