C++ 如何添加到 wstring？
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How do I add to a wstring?
提问by codefrog
This works with std::string
这适用于 std::string
std::string class::something(char* input) {
std::string s(input);
s = "hai! " + s;
return s;
}
But fails if I try the same thing with wstring
但是如果我用 wstring 尝试同样的事情就会失败
std::wstring class::something(wchar_t* input) {
std::wstring s(input);
s = "hai! " + s;
return s;
}
How do I do the same thing with std::wstring?
我如何用 std::wstring 做同样的事情？
回答by Adam Norberg
The problem here is types. A wstring isn't a string, but a quoted string constant is related to it (it is generally a const char*), so
这里的问题是类型。wstring 不是字符串，而是引用的字符串常量与之相关（通常是 a const char*），所以
s = "hai! " + s;
is actually a problem.
其实是个问题。
The value "hai! "is of type const char*, not type const wchar_t*. Since const char*is a basic type, it's searching for a global operator+that operates const char*and wstring, which doesn't exist. It would find one for const wchar_t*and wstring, because std::basic_string<T>, the template underyling type for both stringand wstring(using charand wchar_tas the type parameter, respectively) also creates template methods for operator+ (const T*& s1, const basic_string<T> s2)so that addition can work.
该值"hai! "属于 type const char*，而不是 type const wchar_t*。由于const char*是基本类型，因此它正在搜索不存在的operator+操作const char*and的全局wstring变量。它会发现一个用于const wchar_t*和wstring，因为std::basic_string<T>，模板underyling两种类型string和wstring（使用char和wchar_t分别为类型参数）也创造了模板方法operator+ (const T*& s1, const basic_string<T> s2)，这样除了可以工作。
Therefore, you need to make "hai! " a wstring:
因此，您需要将 "hai!" 变成一个 wstring：
std::wstring class::something(wchar_t* input){
std::wstring s(input);
s = L"hai! " + s;
return s;
}
The Lprefix on a string constant, in Visual C++, defines it to be "long", and therefore a wstring. wstringis actually basic_string<wchar_t>, which is, because of the behavior of C++ templates, a completely different type from basic_string<char>(a std::string), so you can't combine the two.
L在 Visual C++ 中，字符串常量的前缀将其定义为“long”，因此定义为 wstring。wstring实际上basic_string<wchar_t>是 ，也就是说，由于 C++ 模板的行为，与basic_string<char>(a std::string) 的类型完全不同，因此您不能将两者结合起来。
回答by Edward Strange
You use a wide character literal instead of a char character literal by prefixing "hai!" with L.
您可以通过前缀“hai!”来使用宽字符文字而不是字符文字文字。与 L。
回答by Zac Howland
Instead of having 2 versions of the code (1 for wide and 1 for multi-byte):
而不是有 2 个版本的代码（1 个用于宽字节，1 个用于多字节）：
#ifdef _UNICODE
typedef std::wstring tstring;
typedef wchar_t tchar;
# define TEXT(s) L##s
#else
typedef std::string tstring;
typedef char tchar;
# define TEXT(s) s
#endif
tstring class::something(tchar* input)
{
tstring s(input);
s = TEXT("hai! ") + s;
return s;
}
This prevents having to rewriting code when you change your compiler string encoding compiler options.
这可以防止在更改编译器字符串编码编译器选项时必须重写代码。
