为什么 const_cast 不修改调用函数中的值? [英] Why const_cast is not modifying the value in caller function?

问题描述

对于以下代码段，

#include <iostream>
using namespace std;

void fun(const int *p)
{
    int *q = const_cast<int *>(p);
    *q = *q * 10;
    cout<<"q: "<<q<<"	 Value: "<<*q<<endl;
}

int main()
{
    const int a = 10;
    const int *z = &a;
    fun(z);
    cout<<"z: "<<z<<"	"<<"Address of a: "<<&a<<endl;
    cout<<"value at z: "<<*z<<"		 value in a: "<<a<<endl;
}

产生的输出是

q: 0x7fff65910fcc    Value: 100
z: 0x7fff65910fcc   Address of a: 0x7fff65910fcc
value at z: 100      value in a: 10

为什么我尝试在 fun() 中修改 a 的值却没有修改它?

Why the value of a is not modified even though i tried to modify it in fun()?

为什么a和指针z的地址相同，但值不同?

How come the address of a and the pointer z are same but the values are different?

const_cast 是某种未定义的行为吗?

Is it some kind of undefined behavior with const_cast ?

推荐答案

const_cast 是某种未定义的行为吗?

Is it some kind of undefined behavior with const_cast ?

是的，您的程序包含未定义的行为.

Yes, your program contains undefined behavior.

这意味着你不能对其输出有任何期望.原因由 C++11 标准的第 7.1.6.1/4 段给出:

This means that you cannot have any expectation on its output. The reason is given by paragraph 7.1.6.1/4 of the C++11 Standard:

除了声明mutable(7.1.1)的任何类成员都可以修改外，任何修改const的尝试对象在其生命周期 (3.8) 中会导致未定义的行为

Except that any class member declared mutable (7.1.1) can be modified, any attempt to modify a const object during its lifetime (3.8) results in undefined behavior

const_cast 上的第 5.2.11/7 段包含进一步的警告:

Paragraph 5.2.11/7 on const_cast contains a further warning:

[ 注意:根据对象的类型，通过指针、左值或指针进行写操作从 const_cast 产生的数据成员丢弃 const-限定符可能会产生 undefined行为 (7.1.6.1).—尾注 ]

[ Note: Depending on the type of the object, a write operation through the pointer, lvalue or pointer to data member resulting from a const_cast that casts away a const-qualifier may produce undefined behavior (7.1.6.1). —end note ]

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