为什么调用复制构造函数，而不是在此处调用赋值运算符：“Cents cNancy = cMark;” [英] Why Copy Constructor is called, not Assignment operator here : "Cents cNancy = cMark;"

问题描述

Cents cNancy = cMark; // cMark是现有对象



这里调用复制构造函数，而不是赋值运算符？



可以任意一个告诉原因？



谢谢

解决方案

这里假设Cents是一个类，cNancy，cMark实例of Cents class。



Cents cNancy; //这里将调用默认构造函数。

cNancy = cMark; //这里将调用赋值运算符。

Cents cNancy = cMark; //这里新的对象cNancy是通过

调用拷贝构造函数从cMark创建的。

因为这是一个 C ++ 规则。

分数cNancy = cMark; 

与

 Cents cNancy（cMark）;  
 
 
 
即变量初始化：例如，参见Herb Sutter'的政府＃1  - 变量初始化 [ ^ ]。

分配和初始化之间存在差异。



初始化看起来像这样 - int i = 10;

赋值如下所示 - int i; i = 10;



因此规则很简单 - 赋值运算符用于赋值。

Cents cNancy = cMark; //cMark is existing object

Here Copy Constructor is called, not Assignment operator?

Can any one tell why?

Thanks

解决方案

Here assuming that Cents as a class, cNancy, cMark instances of Cents class.

Cents cNancy; // here default constructor will be called.
cNancy = cMark; // here assignment operator will be invoked.
Cents cNancy = cMark;// Here new object cNancy is created from cMark by
calling copy constructor.

Because that is a C++ rule.

Cents cNancy = cMark;

is the same as

Cents cNancy(cMark);

That is variable initialization: see, for instance, Herb Sutter''s GotW #1 - Variable initialization[^].

There is a difference between assignment and initialization.

Initialization looks like this - int i = 10;
Assignment looks like this - int i; i = 10;

So the rule is simple - Assignment operator is used for assignment.

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