运算符=在C ++中重载 [英] Operator= overloading in C++

问题描述

在C ++ Primer的书中，它有一个用于C样式字符数组的代码，并显示如何重载 = / strong>。

  String& String :: operator =（const char * sobj）
 {
 // sobj是空指针，
 if（！sobj）{
 _size = 0; 
 delete [] _string; 
 _string = 0; 
} 
 else {
 _size = strlen（sobj）; 
 delete [] _string; 
 _string = new char [_size + 1]; 
 strcpy（_string，sobj）; 
} 
 return * this; 
} 
  

现在我想知道为什么需要返回一个引用 String& 当下面的代码做同样的工作，没有任何问题：

  void String :: operator =（const char * sobj）
 {
 // sobj是空指针，
 if（！sobj）{
 _size = 0; 
 delete [] _string; 
 _string = 0; 
} 
 else {
 _size = strlen（sobj）; 
 delete [] _string; 
 _string = new char [_size + 1]; 
 strcpy（_string，sobj）; 
} 
 
} 
  

• 请帮助。

解决方案

它支持以下成语：

 字符串a，b; 
 const char * c; 
 
 //将c设置为有趣的东西
 
 a = b = c; 
  

为了工作， b = c 必须返回一个适当的对象或引用赋值给 a ;它实际上是 a =（b = c）根据C ++运算符优先级规则。

指针 this ，你必须写 a = *（b = c）预期意义。

In the book C++ Primer it has a code for C - style character arrays, and shows how to overload the = operator in the Article 15.3 Operator =.

String& String::operator=( const char *sobj )
{
   // sobj is the null pointer,
   if ( ! sobj ) {
      _size = 0;
      delete[] _string;
      _string = 0;
   }
   else {
      _size = strlen( sobj );
      delete[] _string;
      _string = new char[ _size + 1 ];
      strcpy( _string, sobj );
   }
   return *this;
}

Now i would like to know why is there the need to return a reference String & when this code below does the same job, without any problem:

void String::operator=( const char *sobj )
{
   // sobj is the null pointer,
   if ( ! sobj ) {
      _size = 0;
      delete[] _string;
      _string = 0;
   }
   else {
      _size = strlen( sobj );
      delete[] _string;
      _string = new char[ _size + 1 ];
      strcpy( _string, sobj );
   }

}

• please help out.

解决方案

It supports the following idiom:

String a, b;
const char *c;

// set c to something interesting

a = b = c;

For this to work, b = c must return an appropriate object or reference to assign to a; it's actually a = (b = c) according to the C++ operator precedence rules.

If you'd return the pointer this, you'd have to write a = *(b = c), which doesn't convey the intended meaning.

这篇关于运算符=在C ++中重载的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！