C ++复制构造函数语法:和符号是否引用r/l值? [英] C++ copy constructor syntax: Is ampersand reference to r/l values?

问题描述

以下是我的C ++文本的摘录，阐明了使用复制构造函数声明类的语法.

The following is an excerpt from my C++ text, illustrating the syntax for declaring a class with a copy constructor.

class Student {
     int no;
     char* grade;
 public:
     Student();
     Student(int, const char*);
     Student(const Student&);
     ~Student();
     void display() const; 
 };

复制构造函数，如下所示:

The copy constructor, as shown here:

Student(const Student&);

在参数Student后面有一个&符号.

Has an ampersand after the parameter Student.

在C和C ++中，我相信，&字符用作指针的地址"运算符.当然，使用&指针名称之前的字符之前，复制构造函数在其后使用它，因此我认为这是不同的运算符.

In C, and C++ as-well I believe, the ampersand character is used as a 'address of' operator for pointers. Of course, it is standard to use the & character before the pointer name, and the copy constructor uses it after, so I assume this is not the same operator.

我发现的&字符的另一种用法与Rvalues和Lvalues有关，如下所示:

Another use of the ampersand character I found, relates to Rvalues and Lvalues as seen here: http://www.cprogramming.com/c++11/rvalue-references-and-move-semantics-in-c++11.html

我的问题不是关于右值和左值，我只是想知道为什么&字符放置在 参数之后，该参数的名称以及是否/为什么必要.

My question is not about Rvalues and Lvalues, I just want to know why the & character is placed after parameter, and what this is called and if/why it is necessary.

推荐答案

C ++具有C中不存在的引用类型.&用于定义这种类型.

C++ has reference type that does not exist in C. & is used to define such a type.

int i = 10;
int& iref = i;

此处iref是对i的引用.

对i所做的任何更改都可以通过iref看到，而对iref所做的任何更改都可以通过i看到.

Any changes made to i is visible through iref and any changes made to iref is visible through i.

iref = 10; // Same as i = 10;
i = 20;    // Same as iref = 20;

引用可以是左值引用或右值引用.在上面的示例中，iref是左值引用.

The reference can be an lvalue reference or an rvalue reference. In the above example, iref is an lvalue reference.

int&& rref = 10;

此处rref是右值引用.

您可以在 http://en.cppreference.com上了解有关右值引用的更多信息./w/cpp/language/reference .

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