如何更改C++引用所引用的变量？

Question

如果我有这个:

int a = 2;
int b = 4;
int &ref = a;

如何在此代码后ref引用b？

Answer 1

这是不可能的,这是设计的.参考不能反弹.

Answer 2

使用C++ 11,有新的(ish)std :: reference_wrapper.

#include <functional>

int main() {
  int a = 2;
  int b = 4;
  auto ref = std::ref(a);
  //std::reference_wrapper<int> ref = std::ref(a); <- Or with the type specified
  ref = std::ref(b);
}

这对于在容器中存储引用也很有用.

Answer 3

你不能重新分配一个引用,但是如果你正在寻找能够提供相似功能的东西,你可以改为使用指针.

int a = 2;
int b = 4;
int* ptr = &a;  //ptr points to memory location of a.
ptr = &b;       //ptr points to memory location of b now.

您可以使用以下命令获取或设置指针中的值: 

*ptr = 5;     //set
int c = *ptr; //get

Answer 4

您无法重新分配参考.

Answer 5

这不可能以你想要的方式.C++只是不允许你重新绑定引用指向的内容.

但是,如果你想使用技巧,你几乎可以使用新的范围来模拟它(永远不要在真实程序中这样做):

int a = 2;
int b = 4;
int &ref = a;

{
    int& ref = b; // Shadows the original ref so everything inside this { } refers to `ref` as `b` now.
}

Answer 6

从形式上讲,这是不可能的,因为它被设计禁止.任意地说,这是可能的.

引用存储为指针,因此只要您知道如何获取其地址,就可以始终更改它指向的位置.同样,当您无权访问时,您还可以更改const变量,const成员变量甚至私有成员变量的值.

例如,以下代码更改了类A的const私有成员引用:

#include <iostream>
using namespace std;

class A{
private:
    const int &i1;
public:
    A(int &a):i1(a){}
    int geti(){return i1;}
    int *getip(){return (int*)&i1;}
};

int main(int argc, char *argv[]){
    int i=5, j=10;
    A a(i);
    cout << "before change:" << endl;
    cout << "&a.i1=" << a.getip() << " &i=" << &i << " &j="<< &j << endl;
    cout << "i=" << i << " j=" <<j<< " a.i1=" << a.geti() << endl;
    i=6; cout << "setting i to 6" << endl;
    cout << "i=" << i << " j=" <<j<< " a.i1=" << a.geti() << endl;

    *(int**)&a = &j; // the key step that changes A's member reference

    cout << endl << "after change:" << endl;
    cout << "&a.i1=" << a.getip() << " &i=" << &i << " &j="<< &j << endl;
    cout << "i=" << i << " j=" <<j<< " a.i1=" << a.geti() << endl;
    j=11; cout << "setting j to 11" << endl;
    cout << "i=" << i << " j=" <<j<< " a.i1=" << a.geti() << endl;
    return  0;
}

节目输出:

before change:
&a.i1=0x7fff1b624140 &i=0x7fff1b624140 &j=0x7fff1b624150
i=5 j=10 a.i1=5
setting i to 6
i=6 j=10 a.i1=6

after change:
&a.i1=0x7fff1b624150 &i=0x7fff1b624140 &j=0x7fff1b624150
i=6 j=10 a.i1=10
setting j to 11
i=6 j=11 a.i1=11

正如您所看到的,a.i1最初指向i,在更改之后,它指向j.

但是,这样做被认为是危险的,因此不推荐,因为它违背了数据封装和OOP的最初目的.它更像是内存地址黑客攻击.