64位数字和（C）【随意编辑我的英文】

Question

这是我作业的一部分，但我不知道为什么输出不正确。有帮助吗？

/**
 * 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input/output：

 * ./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;
}

当我用gcc编译时，它只显示输出"sum is 310"而不是"sum is 10"？我是C编程新手，还在学习中..

Answer 1

int digitsum(int n)函数错误

你应该在循环中添加每个数字，如下所示 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，如果你的数字是 0，你应该 return 0，否则你用 n % 10 得到你的最后一个数字，然后再次调用你的函数来处理另一个除最后一位以外的数字 digitsum(n / 10):

long long digitsum(long long n) { // I assume n is non-negative
    if (n == 0) return 0;   // terminal case

    return n % 10 + digitsum(n / 10);
}

64位数字和（C）【随意编辑我的英文】

64 Bit digitsum (C) [Feel Free to edit my English]

c

64-bit

sum-of-digits