Question

由于某种原因，这会返回原始值，有人能告诉我原因吗？谢谢！我设置了temp，不确定这是否正确，因为num1和num2需要设置为相等，如果我没有，它们都会保持1个值。

代码：

#include <iostream>
using namespace std;

void swap1(double& num1, double& num2);
void swap2(double& num1, double& num2);

double num1, num2;

int main(){
    cout << "Give me two numbers to swap: ";
    cin >> num1 >> num2;

    cout << "Original: " << num1 << ", " << num2 << endl;
    void swap1(double& num1, double& num2);
    cout << "First swap: " << num1 << ", " << num2 << endl;
    void swap2(double& num1, double& num2);
    cout << "Second swap: " << num1 << ", " << num2 << endl;


}

void swap1(double& num1, double& num2){
    double temp1;
    temp1 = num1;
    num1 = num2;
    num2 = temp1;
}

void swap2 (double& num1, double& num2){
    double temp2;
    temp2 = num1;
    num1 = num2;
    num2 = temp2;
}

输出：抱歉格式错误！

Running /home/ubuntu/workspace/ASSIGN#31.cpp
Give me two numbers to swap: 10 20
Original: 10, 20
First swap: 10, 20
Second swap: 10, 20

Answer 1

cout << "Original: " << num1 << ", " << num2 << endl;
void swap1(double& num1, double& num2);

这不会调用swap1。它是（重新）声明，而不是函数调用。请改用：

swap1(num1, num2);

Answer 2

变化：

 void swap1(double& num1, double& num2);
 void swap2(double& num1, double& num2);

在您的主要功能中：

swap1(num1, num2);
swap2(num1, num2);

它应该可以工作，因为你正在使用一个函数，而不是在你的主函数中声明它。

C ++ SWAP函数什么都不做？

2 个答案: