该解决方案不接受任何浮动

Question

目前在 cs50x 问题集之一 'Cash' 上，这是一个简单的 'ask for how much change is owed, then calculate how many coins are required' 任务，所以不在这里寻求解决方案，但我不明白为什么这行不通.

虽然它确实要求输入，但当我输入诸如 5.96 之类的浮点数时，它只是挂起。没有 returns，没有任何错误。我必须强行关闭它。另一件事是 while 循环在设置为 0 时做同样的事情，这是为了获得硬币的确切数量而设计的做事方式。

我知道这段代码效率低下，而且有更简单的方法来做事。我只是想了解原因，以避免继续犯同样的错误。谢谢

#include <stdio.h>
#include <cs50.h>
#include <math.h>


int main(void)
{
    // changes and other containers
    int p = 1;
    int n = 5;
    int d = 10;
    int q = 25;
    int x = 0;
    float c;
    // get how much change is owed in float
    do
    {
        c = get_float("Change owed: ");
    }
    while (c < 0);
    // int conversion to avoid imprecision
    int a = round(c * 100);
    // 1 because 0 spits out an unknown error
    while (a >= 1)
    {
        // if the converted amount is bigger than a quarter
        if (a >= q)
        {
            // x = number of coins, a = amount left
            x = a / q;
            a = a % q;
        }

        else if (a >= d)
        {
            x = x + a / d;
            a = a - a % d;
        }

        else if (a >= n)
        {
            x = x + a / n;
            a = a - a % n;
        }

        else
        {
            x = x + a / p;
            a = a - a % p;
        }
    }
    printf("%i\n", x);
    printf("%i\n", a);
}

Answer 1

感谢 WhozCraig，我发现我的逻辑有问题。

#include <stdio.h>
#include <cs50.h>
#include <math.h>


int main(void)
{
    // changes and other containers
    int p = 1;
    int n = 5;
    int d = 10;
    int q = 25;
    int x = 0;
    float c;
    // get how much change is owed in float
    do
    {
        c = get_float("Change owed: ");
    }
    while (c < 0);
    // int conversion to avoid imprecision
    int a = round(c * 100);
    // 1 because 0 spits out an unknown error
    while (a >= 1)
    {
        // if the converted amount is bigger than a quarter
        if (a >= q)
        {
            // x = number of coins, a = amount left
            x = a / q;
            a = a % q;
        }

        else if (a >= d)
        {
            x = x + a / d;
            a = a % d;
        }

        else if (a >= n)
        {
            x = x + a / n;
            a = a % n;
        }

        else
        {
            x = x + a / p;
            a = a % p;
        }
    }
    printf("%i\n", x);
    printf("%i\n", a);
}

该解决方案不接受任何浮动

This solution would not accept any floats

c

cs50