什么是常量引用？ （不是常数的引用） [英] What is a constant reference? (not a reference to a constant)

问题描述

一个漂亮的理论问题...为什么常量引用的行为方式与常量指针不同，我可以实际更改他们指向的对象？他们真的看起来像另一个简单的变量声明。为什么我会使用它们？这是一个简单的例子，我运行它编译和运行没有错误：

A pretty theoretical question...Why constant references do not behave the same way as constant pointers and I can actually change the object they are pointing to? They really seem like another plain variable declaration. Why would I ever use them? This is a short example that I run which compiles and runs with no errors:

int main (){
    int i=0;
    int y=1;    
    int&const icr=i;
    icr=y;          // Can change the object it is pointing to so it's not like a const pointer...
    icr=99;         // Can assign another value but the value is not assigned to y...
    int x=9;
    icr=x;
    cout<<"icr: "<<icr<<", y:"<<y<<endl; 
}

推荐答案

X& const x有什么意义吗？

The clearest answer. Does "X& const x" make any sense?

不，这是无稽之谈。

要找到上面的声明的含义，请从右到左阅读：
x是对X的const引用。但这是多余的 - 引用
总是const，在某种意义上，你永远不能重置一个引用
，使它引用一个不同的对象。决不。具有或不具有
const。

To find out what the above declaration means, read it right-to-left: "x is a const reference to a X". But that is redundant — references are always const, in the sense that you can never reseat a reference to make it refer to a different object. Never. With or without the const.

换句话说，X& const x在功能上等同于X& X。
由于在&之后添加const，你就没有什么了，你
不应该添加它：它会混淆人们 - const会使一些
人认为X是const，就好像你说过const X& x。

In other words, "X& const x" is functionally equivalent to "X& x". Since you’re gaining nothing by adding the const after the &, you shouldn’t add it: it will confuse people — the const will make some people think that the X is const, as if you had said "const X& x".

这篇关于什么是常量引用？ （不是常数的引用）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！