c - 将数字分开然后求和的函数

Question

我正在尝试解决一个需要我获取号码​​的问题。使用 % 和/取最右边的数字并对分隔的数字求和。然后判断这个数字是否能被 9 整除。

我创建了一个函数来分隔最右边的数字，然后我尝试获取该数字并通过 while 循环运行它。存在的问题是，当我运行 while 循环时。它会创建无限循环，否则将不会打印输出。

#include <stdio.h>

int loopnum(int n);

int main(void)
{
    int num;
    int sum = 0;
    int d = loopnum(num);

    printf("Enter a number:\n");
    scanf("%d", &num);

    while (loopnum(num) > 0) {
        printf("d = %d", d);
        printf(",sum = %d\n", sum);
    }

    if (num % 9 == 0) {
        printf("n = %d is divisible by 9\n", num);
    }
    else {
        printf("n = %d is not divisible by 9\n", num);
    }
    return 0;
}

int loopnum(int n)
{
    n = n % 10;
    n = n / 10;
    return n;
}

Enter a number:9n = 9 is divisible by 9

这段代码的结果应该是输出 d = "the digital",sum ="digit + sum"且 n = "digit"能被 9 整除。例如，如果我输入 9。输出为 d = 9，sum = 9。

Answer 1

这个函数没有意义。

int loopnum(int n)
{
    n = n % 10;
    n = n / 10;
    return n;
}

例如，对于n等于27，您将得到

n = n % 10;

现在n等于7然后

n = n / 10;

现在n等于0。

因此，对于数字 27，函数返回 0。

而且在循环内部

while (loopnum(num) > 0) {
    printf("d = %d", d);
    printf(",sum = %d\n", sum);
}

num、d 和 sum 均未更改。

定义这样一个函数没有什么意义，除非它本身可以输出中间和和数字。

如果没有该函数，程序可能会如下所示。

#include <stdio.h>

int main(void) 
{
    while ( 1 )
    {
        const unsigned int Base    = 10;
        const unsigned int DIVISOR = 9;

        printf( "Enter a non-negative number (0 - exit): " );

        unsigned int n;

        if ( scanf( "%u", &n ) != 1 || n == 0 ) break;

        unsigned int sum = 0;

        unsigned int tmp = n; 
        do
        {
            unsigned int digit = tmp % Base;
            sum += digit;

            printf( "d = %u, sum = %u\n", digit, sum );
        } while ( tmp /= Base );

        printf( "\nn = %u is %sdivisble by %u.\n\n", 
                 n, 
                 sum % DIVISOR == 0 ? "" : "not ", DIVISOR );

    }

    return 0;
}

它的输出可能看起来像

Enter a non-negative number (0 - exit): 9
d = 9, sum = 9

n = 9 is divisble by 9.

Enter a non-negative number (0 - exit): 123456
d = 6, sum = 6
d = 5, sum = 11
d = 4, sum = 15
d = 3, sum = 18
d = 2, sum = 20
d = 1, sum = 21

n = 123456 is not divisble by 9.

Enter a non-negative number (0 - exit): 0