c++ - 如果类具有引用成员，为什么合成的复制赋值运算符被定义为已删除？

Question

在 C++ Primer，第五版，§13.1.6 中:

The synthesized copy-assignment operator is defined as deleted if a member has a deleted or inaccessible copy-assignment operator, or if the class has a const or reference member.

本章解释:

Although we can assign a new value to a reference, doing so changes the value of the object to which the reference refers. If the copy-assignment operator were synthesized for such classes, the left-hand operand would continue to refer to the same object as it did before the assignment. It would not refer to the same object as the right-hand operand. Because this behavior is unlikely to be desired, the synthesized copy-assignment operator is defined as deleted if the class has a reference member.

复制类会更改引用成员所引用的对象。这不是想要的吗？为什么解释说“不太可能被期望”？

具体来说，

class A {
public:
    A(int &n) : a(n) {}
private:
    int &a;
};

int main() {
    int n = 1;

    A a(n);

    /* Why is this allowed? */
    A b(a);

    /*
    Why is this not allowed?
    error C2280: 'A &A::operator =(const A &)': attempting to reference a deleted function
    */
    b = a;

    return 0;
}

Answer 1

引用一旦创建就无法重新分配。这意味着如果类包含引用成员，则不可能进行正确的赋值运算符。

复制构造函数是另一回事，因为引用可以在对象创建时分配。