如何修复我的 if 语句未被识别为真

Question

我编写了我的程序，该程序应该检查某人是否单身、已婚（分居或共同）以及他们是否是一家之主，以计算他们必须支付的税款。由于某种原因，当我为第一个 scanf 键入 y 时，下一个 if 语句被跳过。请帮忙！这是我的整个代码。我知道它不是最优化的，我可以清理它，但我只是想先弄清楚它有什么问题。

    void question3 (void)
    {
    char single, head, joint;
    int income = 0;
    double tax = 0;

    printf("What is your total income?");
    scanf("%d", &income);

    printf ("Are you single? (y/n) \n");
    scanf (" %c", &single);

    if (single == 'y')
    {
        if (income == 17850)
        {
            tax = income * .15;
        }
        else if (income >= 17850)
        {
            tax = ((.15 * 17850) + (.28 * (income - 17850)));
        }
    }
    else
    {
        if (single != 'y')
        {
        printf("Are you the head of the house hold? y or n\n");
        scanf(" %c", &head);
        }
    }

    if (head == 'y')
    {
        if (income == 23900)
        {
         tax = income * .15;
        }
        else if (income >= 23900)
        {
         tax = ((.15 * 23900) + (.28 * (income - 23900)));
        }
    }
    else if (head != 'y')
    {
        printf("Do you have a joint marriage\n");
        scanf(" %c", &joint);
    }

    if (joint == 'y')
    {
        if (income == 29750)
        {
         tax = income * .15;
        }
        else if (income >= 29750)
        {
        tax = ((.15 * 29750) + (.28 * (income - 29750)));
        }
    }
    else if (joint != 'y')
    {
        if(income == 14875)
        {
         tax = income * .15;
        }
        else if (income >= 14875)
        {
         tax = ((.15 * 14875) + (.28 * (income - 14875)));
        }

    }

    printf("You owe %f dollars in taxes.", tax);
    }

Answer 1

您可以使用函数将其分解，这样对于每个真正的表达式，它都会调用一个函数。在每个功能中，您都要求收入，然后给出结果。打印结果后，它会调用 question3 来重复循环。

对于每一个错误的表达式，你都会给出另一种选择，它要么调用一个函数，要么给出另一种选择。

char single, head, joint;
int income = 0;
double tax = 0;

void Single();
void Head();
void Joint();

void question3(void)
{
    printf("Are you single? (y/n) \n");
    scanf(" %c", &single);

    if (single == 'y')
    {
        Single();
    }
    else if (single == 'n')
    {
        printf("Are you the head of the house hold? y or n\n");
        scanf(" %c", &head);

        if (head == 'y')
        {
            Head();
        }
        else if (head == 'n')
        {
            printf("Do you have a joint marriage\n");
            scanf(" %c", &joint);

            if (joint == 'y')
            {
                Joint();
            }
            else if(joint == 'n')
            {
               question3();
            }
        }
    }
}

void Single()
{
    printf("What is your total income?");
    scanf("%d", &income);

    if (income == 17850)
    {
        tax = income * .15;
    }
    else if (income >= 17850)
    {
        tax = ((.15 * 17850) + (.28 * (income - 17850)));
    }

    printf("You owe %f dollars in taxes.\n\n", tax);

    question3();

    return 0;
}

void Head()
{
    printf("What is your total income?");
    scanf("%d", &income);

    if (income == 23900)
    {
        tax = income * .15;
    }
    else if (income >= 23900)
    {
        tax = ((.15 * 23900) + (.28 * (income - 23900)));
    }

    printf("You owe %f dollars in taxes.\n\n", tax);

    question3();

    return 0;
}

void Joint()
{
    printf("What is your total income?");
    scanf("%d", &income);

    if (income == 29750)
    {
        tax = income * .15;
    }
    else if (income >= 29750)
    {
        tax = ((.15 * 29750) + (.28 * (income - 29750)));
    }

    printf("You owe %f dollars in taxes.\n", tax);

    question3();

    return 0;
}

void main()
{
   question3();

return 0;
}

Answer 2

对于单个字符，您应该使用getchar() 而不是scanf：

single = getchar();

如果你坚持scanf:

• 您应该删除空格（因此scanf("%c", &single)）。
• 您可能应该测试scanf 的返回：它返回它可以解析的 arg 数量（在您的情况下，如果可以，则应该为 1，否则为 0）。

另一方面，这是没用的：

if (single == 'y') {
  ...
} else if (single != 'y') {
  ...
}

如果你不关心单个存在绝对等于 y 或 n，那么你应该像这样写：

if (single == 'y') {
  ...
} else {
  ...
}