Question

这是我的作业的一部分，但我不知道为什么输出不正确。帮助吗？

/**
 * Create a function called digitsum that takes an long integer (64                     bits) and
 * sums the value of each of its digits. i.e. the number 316 should sum to
 * 3+1+6 or 10. Return the sum as the value of the function.
 * Hints:
 * - If the number is negative, make it positive before anything else.
 * - n % 10 will give you the value of the first digit.
 * - n / 10 will shift the integer to the right one digit.
 * - You're done summing when n is zero.
 */

示例输入/输出：

 * ./p2 316
 * num = 316, sum = 10
 * ./p2 -98374534984535
 * num = -98374534984535, sum = 77
 */

#include <stdio.h>
#include <stdlib.h>

int digitsum(int n){ //first try using function
    if (n % 10)
        return digitsum == n % 10;
    else if (n / 10)
        return digitsum == n / 10; 
    else     
        return 0;  //done summing when n is zero.
}

 // Read down in the comments and do **NOT** modify main below.


int main(int argc, char **argv)
{
    int64_t n;

    if (argc < 2) {
        printf("usage: %s <int>\n", argv[0]);
        exit(1);
    }
    n = atoll(argv[1]);

    printf("num = %ld, sum = %d\n", n, digitsum(n));
    return 0;
}

当我使用gcc进行编译时，仅显示输出“ sum is 310”而不是“ sum is 10”吗？我是C语言编程的新手，我还在学习。

Answer 1

int digitsum(int n)的功能错误。

您应该循环添加每个数字，例如以下code：

int digitsum(int64_t n){ //first try using function
    int ret = 0;
    if (n < 0)
        n = -n;
    while (n != 0) {
        ret += n % 10;
        n /= 10;
    }
    return ret;  //done summing when n is zero.
}

Answer 2

考虑递归的基本情况，如果您的数字为0，则应返回0，否则您将使用n % 10获得最后一位，然后再次调用函数以处理除最后一位{ {1}}：

digitsum(n / 10)

64位数字和（C）[随时编辑我的英语]

2 个答案: