C++ 独立引用,为什么?

问题:

I read in some good C++ tutorial that independent references do exist, and act like aliasing. 我读了一些好的C++文章,发现独立引用确实存在,并且很像别名。 But… I wonder what it is made for. 但是,,,我想知道他适用于什么情况。 Why should one want to use aliasing.Besides, some piece of code that is not clear to me: 为什么有人想使用别名呢,另外,下面这段代码我不太清楚。

int a;
int &ref = a; // independent reference
int b = 19;
ref = b;
cout << a << “ ” << ref << “\n”;
ref–;
cout << a << “ ” << ref << “\n”;

First, ref is a ‘reference’ to a.I understand from second line of code that address for ref (hence the ampershead) is 首先,ref是一个对a的引用,从第二行代码那我认为ref的地址也就是a,(括号里面不清楚什么意思。貌似是作者写错了)

a. Then, integer ref is assigned the value of b (19). First cout returns a and ref, both equal to 19. 然后呢,整数ref被b(19)被赋值,第一次输出a和ref,都是19,

Why? Isn’t integer a the address for ref? Then, decrements ref, and last cout gives two times 18. a and ref where decremented. 为什么,难道整数a不是给ref的地址?然后ref自减,最后输出了2个18,a和ref都减少了。

Only strange possible interpretation of mystery: here int&; is a type in itself, ‘independent reference to an integer’, 唯一可能的解释就是:在这里,int& 是一个类型,表示对一个整数的独立引用。

and this type means aliasing. Then whatever you do to ref, the same is done to a. 并且这个类型意味着别名,然后不论你对ref做什么。同样的动作都会对a也做。

Is that right? But why should one need aliasing? 对吗?那为什么要使用别名呢?

答案: You are correct. A reference is similar to a const pointer in many regards (e.g. modifying the value of the reference modifies the original) and you cannot bind the reference to anything else but its initial variable, 你说的很正确,一个引用从很多方面看和一个const 指针是很相似的(比如,对引用的修改都会对被引用对象也修改),并且你不能把对初始变量的绑定重新绑定到其他变量上。

but it cannot be null and the syntax of using it is also different (no need for dereferencing (*, ->)). 但是,引用不能为null,并且使用起来语法也有一点点不同(他不需要使用*和->)

Because it is not a pointer, the best description is alias: think of it as a different name for the same variable. 因为他不是一个指针,最好的一个描述就是别名:认为他是一一个变量的另一个名字

Also, if you pass a reference as a function parameter, no copy of the variable is made (like with a pointer and it pointed-to value), and any change to the reference reflects on the original variable. 而且,如果你在函数参数中传递一个引用,不会产生这个变量的一个副本(就像一个指针,但是他指向一个值),同时,任何对引用的改变都会作用到原始变量上。

There are also special rules for const references that allow a temporary value to live beyond its regular lifetime until the const reference it is bound to goes out of scope. 对于const 引用有很多不同的规则,比如,允许一个临时值存在甚至超过了他正常的生命周期,知道对他的const引用离开正常区域。才会消失。

原文:http://stackoverflow.com/q/10667086/764869

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