在“函数参数"中，通过const类型在派生类中具有函数参数变化的虚拟函数.会打破虚拟机制吗? [英] Virtual functions with varying of function parameters in derived class by const type in "function parameter" will break virtual mechanism?

问题描述

一位C ++专家告诉我们，在派生类中使用const更改函数参数类型将破坏虚拟调用机制.

A C++ guru told that varying the function parameter type with const in derived class, will break the virtual call mechanism.

我尝试了一个简单的程序(原谅非标准代码，纯粹是为了测试而编写的)，但事实证明并非如此. 通过const值更改函数参数不会破坏虚拟机制，

I tried a simple program ( forgive for non standard code, written purely for test) , which prove otherwise. The function parameter change by const value will not break the virtual mechanism,

是否有任何原因&指向这种行为的文档?

Is there any reasons & documentation pointing to this behavior?

关于VS 2012编译器的行为&最新的g ++编译器.

Behavior noted with VS 2012 compiler & latest g++ compiler.

#include <iostream>
using namespace std;


class Base
{
public:
Base(){ cout<<"base"<<endl;}
virtual ~Base(){ cout<<"dest base"<<endl;}
virtual void test(const int x){ cout << "base test"<<"x = " << x<<endl;}
};

class Derived : public Base
{
public:
Derived(){ cout<<"derived"<<endl;}
virtual ~Derived(){ cout<<"dest derived"<<endl;}
virtual void test(int x){ cout << "derived test"<<"x = " << x<<endl;}
};
int main() {

    Base *b = new Derived();
    b->test(10);
    delete b;
    return 0;
}

输出:

base
derived
derived testx = 10
dest derived
dest base

推荐答案

顶级cv限定词不是函数签名的一部分，只是被忽略了.

The top level cv-qualifier is not part of function signature, they are simply ignored.

[dcl.fct]/5

生成参数类型列表后，在形成函数类型时会删除所有修改参数类型的顶级cv限定符.

After producing the list of parameter types, any top-level cv-qualifiers modifying a parameter type are deleted when forming the function type.

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