如何从std :: string_view正确创建std :: string? [英] How to correctly create std::string from a std::string_view?

问题描述

我有一堂课

class Symbol_t {
public:
   Symbol_t( const char* rawName ) {
      memcpy( m_V, rawName, 6 * sizeof( char ) );
   };

   string_view strVw() const {
      return string_view( m_V, 6 );
   };

private:
   char m_V[6];

}; // class Symbol_t

并且有一个我无法修改的lib-func:

and there is a lib-func that I can't modify:

extern bool loadData( const string& strSymbol );

如果有局部变量:

Symbol_t   symbol( "123456" );

当我需要调用loadData时，我不敢这样做:

When I need to call loadData, I dare not do it like this:

loadData( string( symbol.strVw().begin(), symbol.strVw().end() ) );

我必须这样做:

string_view svwSym = symbol.strVw();
loadData( string( svw.begin(), svw.end() ) );

我的问题: 第一种方法正确吗?还是我必须使用第二个?

My question: Is the first method correct? or I must use the second one?

因为我认为在方法1中，传递给std :: string的构造函数的迭代器是两个不同的string_vew对象，并且即使从几乎所有的C ++编译器.

Because I think that in Method 1, the iterators I passed to the constructor of std::string, are of two Different string_vew objects, and theoretically the result is undefined, even though we would get expected result with almost all of the C++ compilers.

任何提示将不胜感激！谢谢.

Any hints will be appreciated! thanks.

推荐答案

无需使用c'tor获取范围. std::string具有根据std::string_view，数字10 在列表中.效果是

There is no need to use the c'tor taking a range. std::string has a constructor that operates in terms of std::string_view, number 10 in the list. The effect of which is

template < class T >
explicit basic_string( const T& t, const Allocator& alloc = Allocator() ); 

就像通过std::basic_string_view<CharT, Traits> sv = t;一样，将t转换为字符串视图sv，然后使用sv的内容来初始化字符串，就像通过basic_string(sv.data(), sv.size(), alloc)一样.仅当std::is_convertible_v<const T&, std::basic_string_view<CharT, Traits>>为true和std::is_convertible_v<const T&, const CharT*>为false时，此重载才参与重载解决方案.

Implicitly converts t to a string view sv as if by std::basic_string_view<CharT, Traits> sv = t;, then initializes the string with the contents of sv, as if by basic_string(sv.data(), sv.size(), alloc). This overload only participates in overload resolution if std::is_convertible_v<const T&, std::basic_string_view<CharT, Traits>> is true and std::is_convertible_v<const T&, const CharT*> is false.

由于两个条件都适用于std::string_view本身，因此我们可以简单地将调用写入loadData:

Since both conditions hold for std::string_view itself, we can write the call to loadData as simply:

loadData( std::string( symbol.strVw() ) );

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