Question

所以，经过对很多网站的大量研究，我终于崩溃了，并且开了帐户。我是新手，也是编程的新手，所以请记住：）

我通过制作一个使用Euclid方法获得最大公约数的程序来制作递归的经典例子。我还需要跟踪函数调用自身的次数/递归发生的次数。

我让程序运行得很完美，但我使用全局（非常糟糕！）变量来保存递归计数。我需要用通过我的gcd函数传递的引用参数替换全局变量，但是我对非常很难理解它是如何工作的。这是我的问题。有人可以告诉我如何使用引用参数删除全局变量吗？非常感谢你！

#include <iostream>
using namespace std;

//Prototype functions
int gcd(int first, int second);

//Global varibles
//Depth will count how many times the recursive function runs.
int depth = 0;

int main ()
{
    //variables
    int firstNum, secondNum;
    do
    {
        cout << "Welcome to Euclid's method for finding the GCD." << endl << "To quit, enter 0 for either number." << endl;
        cout << "Enter the first number:";
        cin >> firstNum;
        cout << "Enter the second number:";
        cin >> secondNum;
        //The program quits if the user enters 0. If this happens, the program won't bother with running the recursive function.
        if (firstNum != 0 && secondNum != 0)
        {
            cout << "The GCD of " << firstNum << " and " <<secondNum << " is ";
            cout << gcd(firstNum, secondNum) << "." << endl;
            cout << "The recursive calculation required a depth of " << depth <<"." << endl;
        }
    }
    while (firstNum != 0 && secondNum != 0); //Program quits on 0 being entered
    return (0);
}
//Recursive function. It will call itself until second number is 0.
int gcd(int first, int second)
    {
        if(second != 0) 
        {
            depth++;
            return gcd(second, first % second);
        }
        else
            return first;
    }

Answer 1

尝试通过引用传递该参数。

int main ()
{
    //variables
    int firstNum, secondNum;
    int depth = 0;
    do
    {
        cout << "Welcome to Euclid's method for finding the GCD." << endl << "To quit, enter 0 for either number." << endl;
        cout << "Enter the first number:";
        cin >> firstNum;
        cout << "Enter the second number:";
        cin >> secondNum;
        //The program quits if the user enters 0. If this happens, the program won't bother with running the recursive function.
        if (firstNum != 0 && secondNum != 0)
        {
            cout << "The GCD of " << firstNum << " and " <<secondNum << " is ";
            cout << gcd(firstNum, secondNum, depth) << "." << endl;
            cout << "The recursive calculation required a depth of " << depth <<"." << endl;
        }
    }
    while (firstNum != 0 && secondNum != 0); //Program quits on 0 being entered
    return (0);
}
//Recursive function. It will call itself until second number is 0.
int gcd(int first, int second, int &depth)
    {
        if(second != 0) 
        {
            depth++;
            return gcd(second, first % second, depth);
        }
        else
            return first;
    }

Answer 2

将参考参数添加到函数定义中。

int gcd(int first, int second, int& depth)
    {
        if(second != 0) 
        {
            depth++;
            return gcd(second, first % second, depth);
        }
        else
            return first;
    }

然后在depth中将main()声明为局部变量，将其初始化为0，然后致电gcd(firstNum, secondNum, depth)。

Answer 3

int gcd(int first, int second, int& depth)
{
    if(second != 0) 
    {
        ++depth;
        return gcd(second, first % second, depth);
    }
    else
    {
        return first;
    }
}

你必须把它称之为有点不同。创建新的int，将其初始化为0并作为第三个参数传递。完成此操作后，它将保持深度。

Answer 4

将原型参数的深度添加到（注意参考符号），不需要将其命名为深度。

int gcd(int first, int second, int &depth);

删除深度的全局定义。

为局部变量添加深度

int firstNum, secondNum, depth=0;

为第一次调用添加深度（通过引用传递）

std::cout << gcd(firstNum, secondNum, depth) << "." << std::endl;

为功能定义添加深度

int gcd(int first, int second, int &depth)

为递归调用添加深度

return gcd(second, first % second, depth);

将全局变量更改为参考参数C ++

4 个答案:

将全局变量更改为参考参数C​​ ++

4 个答案:

将全局变量更改为参考参数C ++