c++中关于const_cast的问题

2022-01-05 00:00:00 function reference casting c++

全部:这是引自 Effective C++ 3rd edition

all: this is quoted from Effective C++ 3rd editiion

const_cast 通常用于抛弃对象的常量性.这是唯一可以做到这一点的 C++ 风格的强制转换.

const_cast is typically used to cast away the constness of objects. It is the only C++-style cast that can do this.

我的问题是 const_cast 可以为非常量对象添加常量性吗?其实我写了一个小程序来验证我的想法.

My question is can const_cast add constness to a non-const object? Actually i wrote a small programme trying to approve my thought.

class ConstTest
{
 public:

 void test() {
    printf("calling non-const version test const function 
");
}

 void test() const{
    printf("calling const version test const function 
");

} 

};
 int main(int argc,char* argv){
 ConstTest ct;
 const ConstTest cct;
 cct.test();
 const_cast<const ConstTest&>(ct).test();//it is wrong to write this statement without the '&',why

}

省略&"导致以下错误:

Omitting the '&' incurs error below:

错误 C2440:const_cast":无法从ConstTest"转换为const ConstTest"

error C2440: 'const_cast' : cannot convert from 'ConstTest' to 'const ConstTest'

说明const_cast可以增加constness，但是好像必须强制转换为一个对象引用，这个引用有什么魔力?

It shows that const_cast can add constness,but seems like you have to cast to a object reference, what is the magic of this reference ?

推荐答案

你不需要const_cast来添加constness:

You don't need const_cast to add constness:

class C;
C c;
C const& const_c = c;

反过来需要一个const_cast

const C const_c;
C& c = const_cast<C&>(const_c);

但是如果您尝试在 c 上使用非常量操作，则行为未定义.

but behavior is undefined if you try to use non-const operations on c.

顺便说一句，如果不使用引用，则要制作对象的副本:

By the way, if you don't use a reference, a copy of the object is to be made:

C d = const_c; // Copies const_c

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