const_cast的奇怪行为

问题描述

考虑以下代码：

我声明一个新的引用端，通过const_cast将其赋值给a。然后，我只需增加参考值即可打印地址和值。

I declare a new reference end assign it to value a via const_cast. Then I just increase the reference value print the addresses and values.

#include <iostream>
using namespace std;
int main()
{
    const int a = 7;
    int &b = const_cast<int&>(a);
    ++b;
    cout<<"Addresses  "<<&a<<" "<<&b<<endl;
    cout<<"Values  "<<a<<" "<<b<<endl;
}

//output
Addresses  0x7fff11f8e30c 0x7fff11f8e30c
Values  7 8

我如何在同一个地址中拥有2个不同的值？

How can i have 2 different values in the same address??

推荐答案

修改常量对象提供未定义的行为，因此您的程序（原则上）可以执行任何操作。

Modifying a constant object gives undefined behaviour, so your program could (in principle) do anything.

保持此行为未定义的原因之一是允许优化用其常量替换常量值（因为您已经声明了值永远不会改变）。这就是这里发生的情况： a 在编译时被替换为值 7 ，因此无论什么值都将保留该值您尝试在运行时执行此操作。

One reason for leaving this behaviour undefined is to allow the optimisation of replacing a constant variable with its value (since you've stated that the value can never change). That's what is happening here: a is replaced with the value 7 at compile time, and so will keep that value whatever you try to do to it at run time.

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