为什么我的代码不输出(仅)数字?

时间:2015-10-09 07:36:42

标签: c++ function output

练习提示代码:编写一个程序,告诉你给出的任何金额从1美分到99美分的硬币。使用25美分(四分之一),10美分(一分钱)和1美分(便士)的硬币面额。不要使用镍和半美元硬币。您的程序将使用以下功能(以及其他功能): void compute_coins(int coin_value,int& num,int& amount_left);

#include <iostream>
#include <string>
using namespace std;

void prompt(int *amount_left);
void remaining_change(int *amount_left, int coin_value);
void compute_coins(int coin_value, int *num, int *amount_left);
void output(string coin_name, int *num);



int main() {
    int change = 0, num = 0, amount_left = 0;
    const int quarter = 25, dime = 10, penny = 1;
    string q = "quarter(s)", d = "dime(s)", p = "penny(s)"; 

    prompt(&change);
    compute_coins(quarter, &num, &amount_left);
    remaining_change(&amount_left, quarter);
    output(q, &num);

    compute_coins(dime, &num, &amount_left);
    remaining_change(&amount_left, dime);
    output(d, &num);

    compute_coins(penny, &num, &amount_left);
    output(p, &num);

}

void prompt(int *change)
{
  cout << "How much change is there? ";
  cin >> *change;
  cout << "You entered " << change << endl;
  cout << "That is equal to: ";
}

void remaining_change(int *amount_left, int coin_value)
{
    *amount_left = (*amount_left % coin_value);
}
void compute_coins(int coin_value, int *num, int *amount_left)
{
   *num = *amount_left / coin_value; 
}

void output(string coin_name,int *num)
{
    cout << num << " " << coin_name << ", ";
}

1 个答案:

答案 0 :(得分:1)

您正在输出指针的值,而不是指向的对象的值。

简单的解决方法是首先取消引用指针:

cout << "You entered " << *change << endl;
//                        ^

cout << *num << " " << coin_name << ", ";
//      ^

但是,我建议不要使用指针来完成这样的事情。对于内置类型,您应该在想要更新变量时使用引用,否则应该使用值。

我个人也不会从函数内部更新这些变量,我会执行必要的输入或计算并返回一个要分配的值。