Question

使用命令：g++ -o test.exe test.cpp

使用GCC 4.8.1（MinGW）进行编译

TEST.CPP ：

#include <iostream>

int a = 1, b = 2;

void F()
{
    int &r = a;
    std::cout << r;
    r = b;
    std::cout << r;
}

int main ( int, char** )
{
    F();
    F();
    return 0;
}

预期输出为： 1212 ，但实际为： 1222

当r被定义为指针时，它按预期工作。

void F()
{
    int *r = &a;
    std::cout << *r;
    r = &b;
    std::cout << *r;
}

有人知道“帐篷里的钥匙”在哪里吗？

Answer 1

在带指针的示例中：

r = &b;

这会将指针r重新置于指向b的位置。在带引用的示例中：

r = b;

您可能认为这与指针的示例类似，但事实并非如此。引用不像指针那样可以重新定位。一旦它们被初始化，它们就成为它们引用的东西的别名。所以上面的行与此完全相同：

a = b;

也就是说，它会将b中存储的值复制到a。

参考示例可以这样写：

void F()
{
    std::cout << a;
    a = b;
    std::cout << a;
}

虽然指针示例可以这样写：

void F()
{
    std::cout << a;    
    std::cout << b;
}

Answer 2

您的“问题”如下：

int &r = a;

这将创建一个指向a的所谓引用。虽然引用在访问它们时不需要额外的逻辑（即解除引用;与指针相比），但它们仍然在幕后工作方式相同。

所以一旦你点击

r = b;

您实际上将使用该引用并更新a的值。在这种情况下，r永远不会拥有它自己的价值。它始终指向a并且基本上是别名。

在你的第二个例子中，这是不同的。这条线

int *r = &a;

创建一个指向a和

的指针

r = &b;

将重新分配该地址，使指针现在指向b。

简而言之，当您将一些其他值（地址）指定给指针时，您更改指针，它将指向其他位置。但是，在为参考分配新值时，实际上是在更新值，其中引用，而非参考本身。

Answer 3

//global variables! remember that this will be like a state
//and will be freed on (data?BSS?stack?) only after end of program.
int a = 1, b = 2;

void F()
{
    //You are referencing here,
    //meaning, once you change the referee ( r )
    //the reference will changed permanently.
    //Permanently because you are referring to a global variable.
    int &r = a;        

    std::cout << r;

    //You have changed the value of ( r )
    //forever!

    r = b;   //I think you are assuming the you have changed the reference address here?
    std::cout << r;
}

void F()
{
    //You have declared and initialized a pointer
    //that points to global variable.
    //w/c means you can also change its value forever!
    int *r = &a;
    std::cout << *r;

    //You have changed the ( r ) address pointing to, not its value!
    r = &b;
    std::cout << *r;
}

Answer 4

Your code is absolutely fine and it's working according to your instruction.
Suppose a variable has 1000 address b variable 1004 address, now your first line is  
         int &r = a;
that means r has 1000 address  and you did cout<< r //print 1
In second statement  r = b that means 1000 address hold value of b means 2;
Now , your scenario has been changed 1000---> 2 and 1004--->2 

and again you are calling F();
that's why compiler print twice 2 value. . There is a concept referencing in c++. Hope it  
would be useful for u

隐藏的g ++引用错误

4 个答案: