如何C ++隐式转换c样式字符串为字符串对象？ [英] how C++ Implicitly convert c style string to a string object?

问题描述

  string s =abc; 
  

上述语句将首先调用 string（const char * s） code>构造函数，然后根据调用复制构造函数是C ++中字符串初始化的区别吗？
这里是问题：C ++如何知道应该调用 string（const char * s）将字符串abc转换为临时字符串对象？ p>



注意：
复制初始化时不会调用复制构造函数。

解决方案

使用语法初始化对象

  string s =abc 
  

称为复制初始化。

有几种情况是合法的初始化。在所有情况下，RHS必须可转换为 string 才能正常工作。

一个字符串文字可以转换为 string 是通过 string 的构造函数获取 char const * 。这称为用户定义的转化。

string s = "abc";

The above statement will first invoke string ( const char * s ) constructor, then invoke copy constructor according to What are the differences in string initialization in C++? . Here is the question: how C++ know it should invoke string ( const char * s ) to convert literal string "abc" to a temporary string object?

Note: copy constructor won't be invoked in copy initialization.

解决方案

Initializing an object by using the syntax

string s = "abc";

is called copy initialization.

There are several scenarios where that is legal initialization. In all cases the RHS must be convertible to a string for it to work.

One way a string literal can be converted to a string is through the constructor of string that takes a char const*. That is called a user defined conversion.

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