C中的百分比计算问题

Question

我正在制作一个程序，根据条件计算有多少及格分数。但是，当我尝试计算及格分数的百分比时，即 if (passGrade >= 0) { printf("Percent of passing grades are %d! \n", (inputGrade/passGrade)*100); return 0; 程序没有显示结果，而是显示为零，我确实调试了程序，并且 passGrade 是多少我在这一行中输入的总成绩显示了它的值 printf("You have entered %d passing grades! \n", passGrade); 但是当涉及到下一行时，它只显示了一个 -1。我确实在网上查找问题，但没有任何显示。

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
int main()
{

    int inputGrade, passGrade, percentGradePass;

    inputGrade = 0;
    passGrade = 0;

    /*  printf("Please type a grade (-1) to exit: ");
      scanf("%i", &inputGrade);
      printf("You typed: %i \n", inputGrade); */



    while (inputGrade != -1)
    {

        printf("Please type a grade (-1) to exit: ");
        scanf("%d", &inputGrade);
        printf("You typed: %d \n", inputGrade);

        if (inputGrade == -1)
        {
            inputGrade = -1;
        }
        else if (inputGrade >= 70 && inputGrade <= 100) {

            passGrade = passGrade + 1;

        }
        else {
            inputGrade = 0;
        }
    }


    printf("You have entered %d passing grades! \n", passGrade);
    if (passGrade >= 0) {
        printf("Percent of passing grades are %d! \n", (inputGrade/passGrade)*100);
        return 0;
   }
       

}

不胜感激。

Answer 1

'the program does not show the result' 是什么意思？

当您的程序达到“(inputGrade/passGrade)*100)”时

1/ 你应该检查 'passGrade' 是否为 0，因为如果它是 '0' 就会出现 run-time 错误。

2/ 而且，由于 while 循环在 'inputGrade == -1' 时结束，所以您需要将其转换为 'plus'，然后再除以

3/ 最后，因为 'scanf("%d", &inputGrade);' 表示 inputGrade = 'input' 所以，你不需要在你的 'if' 案例中写 'inputGrade = -1;'

Answer 2

您的代码存在许多问题。有些问题很关键，而另一些则代表“strange/unnecessary”代码。

内联注释示例：

while (inputGrade != -1)
{

    printf("Please type a grade (-1) to exit: ");
    scanf("%d", &inputGrade);   // ALWAYS, ALWAYS... check the return value of scanf
    printf("You typed: %d \n", inputGrade);

    if (inputGrade == -1)
    {
        inputGrade = -1;  // Unnecessary! inputGrade is already -1 so no need to 
                          // assign it again
    }
    else if (inputGrade >= 70 && inputGrade <= 100) {

        passGrade = passGrade + 1;

    }
    else {
        inputGrade = 0;  // Unnecessary! inputGrade will get a new value from scanf 
                         // in the next loop
    }
}

所以你的循环可以简单地是：

while (inputGrade != -1)
{

    printf("Please type a grade (-1) to exit: ");
    if (scanf("%d", &inputGrade) != 1)
    {
        // Invalid input, terminate program
        exit(1);
    }
    
    printf("You typed: %d \n", inputGrade);

    if (inputGrade >= 70 && inputGrade <= 100) 
    {
        passGrade = passGrade + 1;
    }
}

另一个潜在的问题是：

(inputGrade/passGrade)*100

因为两个变量的类型都是int，除法的结果将被截断为最接近的整数，例如6/10 将导致 0 而不是 0.6，整个表达式将因此变为 0。为避免这种情况，您可以先进行乘法（即 (100 *inputGrade)/passGrade）或使用浮点类型而不是整数类型。

也就是说，你的算法似乎存在基本的误解。你说：

I try calculating the percentage of passing scores

计算你想要的百分比

100 * NumberOfPassingGrades / TotalNumberOfGrades

但这不是你在做什么。不是在计算中使用inputGrade，而是需要计算成绩的数量并在计算中使用它。

类似于：

int totalGrade = 0;
while (1)
{

    printf("Please type a grade (-1) to exit: ");
    if (scanf("%d", &inputGrade) != 1)
    {
        // Invalid input, terminate program
        exit(1);
    }

    if (inputGrade != -1)
    {
        // Stop the loop
        break;
    }
    
    printf("You typed: %d \n", inputGrade);

    if (inputGrade >= 70 && inputGrade <= 100) 
    {
        passGrade = passGrade + 1;
    }
    totalGrade = totalGrade  + 1;
}

printf("You have entered %d passing grades! \n", passGrade);
printf("You have entered %d grades in total \n", totalGrade);
if (totalGrade > 0) 
{
    printf("Percent of passing grades are %d! \n", (100 * passGrade)/totalGrade);
}

C中的百分比计算问题

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