关于C ++中的指针和引用的困惑 [英] Confusion about pointers and references in C++

问题描述

我有一堆这样的代码:

#include <iostream>
using namespace std;

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

int main() {

    int a;
    int b;
    a = 7;
    b = 5;
    swap(a, b);
    cout << a << b;

    return 0;
}

此代码执行交换过程，就像我要交换2个数字一样

This code does the swapping process as what I exactly wanted to swap 2 numbers

但是当我想要从用户那里获得两个数字时，如下所示；

But when I want two numbers from the user as follows;

int a;
int b;
cin >> a;
cin >> b;
swap(a, b);
cout << a << b;

编译器给我一个关于整数到*整数错误的错误，这是预期的.尽管我没有在&运算符中使用该方法，但是为什么第一个代码可以正确地进行交换?

The compiler gives me an error about integer to *integer error which is as expected. Why does the first code do the right swapping although I didn't use the method with & operator?

推荐答案

在第一个示例中，由于您的using namespace std而调用std::swap. 第二个示例与第一个示例完全相同，因此您可能没有用.

In the first example, std::swap is called, because of your using namespace std. The second example is exactly the same as the first one, so you might have no using.

无论如何，如果将函数重命名为my_swap或类似的名称(并更改每次出现的值)，则第一个代码将无法正常工作.或者，删除using namespace std并显式调用std::cin和std::cout.我建议第二种选择.

Anyway, if you rename your function to my_swap or something like that (and change every occurence), then the first code shouldn't work, as expected. Or, remove the using namespace std and call std::cin and std::cout explicitly. I would recommend the second option.

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