c - 为什么我的程序在某些测试中可以正常工作，而在其他测试中却不能正常工作？

Question

这个问题已经有答案了:

Is floating point math broken? (33 个回答)

已关闭 5 年前。

这是我的测试的输出:

:) greedy exists

:) greedy compiles

:( input of 0.41 yields output of 4
    expected "4\n", not "3\n"

:( input of 0.01 yields output of 1
    expected "1\n", not "0\n"

:) input of 0.15 yields output of 2

:) input of 1.6 yields output of 7

:) input of 23 yields output of 92

:) input of 4.2 yields output of 18

:) rejects a negative input like -.1

:) rejects a non-numeric input of "foo"

:) rejects a non-numeric input of ""

这是代码:

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

void count_coins();

int coin = 0;
float change = 0;

int main(void)
{
    do
    {
    change = get_float("How much change is owed? ");
    }
    while (change < 0);

    count_coins();

    printf("%i\n", coin);
}

void count_coins()
{
    while (change > 0.24)
    {
        coin++;
        change -= 0.25;
        // printf("%.2f\n", change);
    }

    while (change > 0.09)
    {
        coin++;
        change -= 0.10;
        // printf("%.2f\n", change);
    }

    while(change > 0.04)
    {
         coin++;
         change -= 0.05;
         // printf("%.2f\n", change);
    }

    while (change >= 0.01)
    {
        coin++;
        change -= 0.01;
        // printf("%.2f\n", change);
    }
}

Answer 1

正如 Havenard 已经写的那样，问题在于 change 存储为 float。对于这样的程序，更改必须存储为整数值。

这是您的代码，使用 int 而不是 float:

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

void count_coins (void);

int coin = 0;
int change = 0;

int main (void) {
    do {
        change = 41; // insert a get integer function here
    } while (change < 0);

    count_coins();

    printf("%i\n", coin);
}

void count_coins (void) {
    while (change >= 25) {
        coin++;
        change -= 25;
    }

    while (change >= 10) {
        coin++;
        change -= 10;
    }

    while(change >= 5) {
        coin++;
        change -= 5;
    }

    while (change >= 1) {
        coin++;
        change -= 1;
    }
}