問題描述
我有一堆這樣的代碼:
#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;
}
這段代碼按照我真正想要交換 2 個數字的方式執行交換過程，但是當我想要用戶提供兩個數字時，如下所示；
This code does the swapping process as what I exactly wanted to swap 2 numbers, 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;
編譯器給了我一個關于 int 到 int* 錯誤的錯誤，這是預期的.盡管我沒有使用帶有 & 運算符的方法，為什么第一個代碼會進行正確的交換?
the compiler gives me an error about int to int* error which is as expected. Why does the first code do the right swapping although I didn't use the method with & operator?
推薦答案
在第一個例子中，std::swap 被調用，因為你的using namespace std.第二個例子和第一個完全一樣，所以你可能沒有用.
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.
無論如何，如果您將函數重命名為 my_swap 或類似名稱(并更改每次出現的次數)，那么第一個代碼不應按預期工作.或者，刪除 using namespace std 并顯式調用 std::cin 和 std::cout.我會推薦第二個選項.
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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