向前声明的std :: wstring [英] Forward declaration of std::wstring

问题描述

// This is a header file.

class MyClass; // It can be forward declared because the function uses reference.
// However, how can I do forward declaraion about std::wstring?
// class std::wstring; doesn't work.
VOID Boo(const MyClass& c);
VOID Foo(const std::wstring& s);

推荐答案

#include< string> ，您几乎没有选择。

原因是 wstring 在命名空间 std 中定义，并且被typedef定义到 std :: basic_string< wchar_t& code>。更精确地， std :: wstring 是 std :: basic_string< wchar_t，std :: char_traits< wchar_t> > 。这意味着为了转发声明 std :: wstring ，你必须转发声明 std :: char_traits<> 和 std :: basic_string<> 在命名空间 std 内。因为（除了少数例外）标准禁止添加定义或声明到命名空间 std （17.4.3.1/1）最终你不能转发声明任何标准模板或以符合标准的方式输入。具体来说，这意味着你不能转发声明 std :: wstring 。

The reason is that wstring is defined in namespace std and is typedef'd to std::basic_string<wchar_t>. More elaborately, std::wstring is std::basic_string<wchar_t, std::char_traits<wchar_t> >. This means that in order to forward-declare std::wstring you'd have to forward-declare std::char_traits<> and std::basic_string<> inside namespace std. Because (apart from a few exceptions) the standard forbids adding definitions or declarations to namespace std (17.4.3.1/1) ultimately you can't forward-declare any standard template or type in a standard-conforming way. Specifically, this means you can't forward-declare std::wstring.

是的，我们都同意将< stringfwd> 头，比如< iosfwd> >< iostream> 。但没有。 < string> 也不像编辑< iostream> 您有两种选择：＃include< string> 或使用不透明指针。

And yes, we all agree it would be convenient to have a <stringfwd> header, like <iosfwd> for <iostream>. But there isn't. <string> is also not nearly as hardcore to compile as <iostream>, but nevertheless. You have two choices: #include<string> or use an opaque pointer.

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