为什么我不能对普通变量进行多态?-C/C++问题

Why can#39;t I do polymorphism with normal variables?(为什么我不能对普通变量进行多态?)

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问题描述

我是一名 Java 程序员，最近开始学习 C++.我被某些事情搞糊涂了.

I'm a Java programmer and recently started studying C++. I'm confused by something.

我知道在 C++ 中，要实现多态行为，您必须使用指针或引用.例如，考虑一个带有实现方法 getArea() 的 Shape 类.它有几个子类，每个子类都以不同的方式覆盖 getArea().比考虑以下功能:

I understand that in C++, to achieve polymorphic behavior you have to use either pointers or references. For example, consider a class Shape with an implemented method getArea(). It has several subclasses, each overriding getArea() differently. Than consider the following function:

void printArea(Shape* shape){
    cout << shape->getArea();
}

函数根据指针指向的具体Shape调用正确的getArea()实现.

The function calls the correct getArea() implementation, based on the concrete Shape the pointer points to.

效果相同:

void printArea(Shape& shape){
    cout << shape.getArea();
}

但是，以下方法不能多态地工作:

However, the following method does not work polymorphicaly:

void printArea(Shape shape){
    cout << shape.getArea();
}

不管函数中传入什么样的Shape，都会调用相同的getArea()实现:Shape<中的默认实现/代码>.


Doesn't matter what concrete kind of Shape is passed in the function, the same getArea() implementation is called: the default one in Shape.
我想了解这背后的技术推理.为什么多态适用于指针和引用，而不适用于普通变量?(而且我想这不仅适用于函数参数，而且适用于任何东西.
I want to understand the technical reasoning behind this. Why does polymorphism work with pointers and references, but not with normal variables? (And I suppose this is true not only for function parameters, but for anything).
请解释这种行为的技术原因，以帮助我理解.
Please explain the technical reasons for this behavior, to help me understand.
推荐答案
答案是复制语义.
当您在 C++ 中按值传递对象时，例如printArea(Shape shape) 一个副本是你传递的对象.如果将派生类传递给该函数，则复制的只是基类 Shape.仔细想想，编译器不可能做任何其他事情.
When you pass an object by value in C++, e.g. printArea(Shape shape) a copy is made of the object you pass. And if you pass a derived class to this function, all that's copied is the base class Shape. If you think about it, there's no way the compiler could do anything else.
Shape shapeCopy = circle;

shapeCopy 被声明为 Shape，而不是 Circle，所以编译器所能做的就是构造一个  的副本对象的形状部分.
shapeCopy was declared as a Shape, not a Circle, so all the compiler can do is construct a copy of the Shape part of the object.

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