从 C++ 中的 const 成员函数修改引用成员

2022-01-05 00:00:00 reference constants c++

我正在研究我的代码的常量正确性，只是想知道为什么这段代码会编译:

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，尽管它说是 const.当我分配给 y 时，如何确保我的编译器抱怨?(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)

非常感谢！

推荐答案

因为您没有更改 X 中的任何变量.实际上，您正在更改 _y ，它是您班级的局外人.别忘了:

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，而不是它们本身的引用.仅在初始化时考虑引用.

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.

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