为什么是复制构造函数的参数const？ [英] Why is the copy-constructor argument const?

问题描述

 Vector(const Vector& other) // Copy constructor 
 {
    x = other.x;
    y = other.y;

为什么参数是const？

Why is the argument a const?

推荐答案

你已经得到了答案，提到确保ctor不能改变被复制的东西 - 他们是对的，把const有效果。

You've gotten answers that mention ensuring that the ctor can't change what's being copied -- and they're right, putting the const there does have that effect.

更重要的是，一个临时对象不能绑定到一个非const引用。复制ctor必须引用一个const对象，以便能够复制临时对象。

More important, however, is that a temporary object cannot bind to a non-const reference. The copy ctor must take a reference to a const object to be able to make copies of temporary objects.

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