在C ++中从const成员函数修改引用成员 [英] Modifying reference member from const member function in C++

问题描述

我正在处理我的代码的const正确性，只是想知道为什么这个代码编译：

I am working on const-correctness of my code and just wondered why this code compiles:

class X
{
    int x;
    int& y;
public:
    X(int& _y):y(_y)
    {
    }
void f(int& newY) const
    {
        //x = 3; would not work, that's fine
        y = newY; //does compile. Why?
    }
};

int main(int argc, char **argv) 
{
    int i1=0, i2=0;
    X myX(i1);
    myX.f(i2);
...
}

据我所知，f正在改变对象myX，虽然它说是const。我怎么能确保我的编译器抱怨，当我分配到y？ （Visual C ++ 2008）

As far as I understand, f() is changing the object myX, although it says to be const. How can I ensure my compiler complains when I do assign to y? (Visual C++ 2008)

感谢很多！

推荐答案

您不会更改 X 中的任何变量。其实，你正在改变 _y 这是一个局外人对你的类。不要忘记：

Because you are not changing any variable in X. Actually, you are changing _y which is an outsider with respect to your class. Don't forget that:

y = newY;

正在将 newY 的值分配给 y 指向的变量，但不是引用它们自己。只有在初始化时，才考虑引用。

Is assigning the value of newY to the variable pointed by y, but not the references them selves. Only on initialization the references are considered.

这篇关于在C ++中从const成员函数修改引用成员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！