通过引用问题C++传递

Question

有人可以帮助解释为什么当我乘以2个数字时它会返回0,但它有3个和4个数字吗？

我为一堂课写了这篇文章,但我不知道它有什么不对,谢谢.

功能是使用重载函数乘以2,3或4个数字,并通过引用传递产品.

#include <iostream>
using namespace std;

void mult(int& product, int n1, int n2);
void mult(int& product, int n1, int n2, int n3);
void mult(int& product, int n1, int n2, int n3, int n4);

int main() {
    int product, n1, n2, n3, n4;
    char ans;

    do {
        product = 0;
        n1 = 0;
        n2 = 0;
        n3 = 0;
        n4 = 0;
        cout << "Enter 2-4 numbers to multiply\n";
        cout << "First number: ";
        cin >> n1;
        cout << "Second number: ";
        cin >> n2;
        cout << "Enter a 3rd number? (y/n):";
        cin >> ans;

        if (ans == 'y') {
            cout << "Third number: ";
            cin >> n3;
            cout << "Enter a 4th number? (y/n):";
            cin >> ans;
        } else {
            mult(product, n1, n2);
        }

        if (ans == 'y') {
            cout << "Fourth number: ";
            cin >> n4;
            mult(product, n1, n2, n3, n4);
        } else {
            mult(product, n1, n2, n3);
        }

        cout << "The product is " << product << endl << n1 << n2 << n3 << n4;
        cout << "Would you like to calculate another? (y/n):";
        cin >> ans;

    } while (ans == 'y');
}

定义

void mult(int& product, int n1, int n2)
{
    product = (n1 * n2);
    cout << product;
}
void mult(int& product, int n1, int n2, int n3)
{
    product = (n1 * n2 * n3);
}
void mult(int& product, int n1, int n2, int n3, int n4)
{
    product = (n1 * n2 * n3 * n4);
}

Answer 1

这是因为您的控制结构执行语句

else{mult(product,n1,n2,n3);}

即使你只打算使用mult(product,n1,n2).只有两个数字,n3将为0.因此结果也为零.

您可以通过重组它来解决它:

   cout << "Enter a 3rd number? (y/n):";
    cin >> ans;

    if (ans == 'y') {
        cout << "Third number: ";
        cin >> n3;
        cout << "Enter a 4th number? (y/n):";
        cin >> ans;
        if (ans == 'y') {
            cout << "Fourth number: ";
            cin >> n4;
            mult(product, n1, n2, n3, n4);
        } else {
            // Three numbers
            mult(product, n1, n2, n3);
        }
    } else {
        // Two numbers
        mult(product, n1, n2);
    }