C-从左到右添加2个大数字

Question

是否可以添加2个大数字而不反转数组？我必须使用此功能声明：

int add(const char* n1, const char* n2, char** sum);

我无法反向数组，因为它是 cosnt char* ：(

Answer 1

我认为这是一个挑战 - 以下代码从左到右添加了两个非负号字符串。它只需从较长的时间开始，直到数字对齐，然后添加字符。如果有携带，它会向右传播，以左修复已经求和的数字：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int add(const char *n1, const char *n2, char **sum) {
    const char *longer = n1, *shorter = n2; // ['1', '2', '3'] + ['8', '9']
    size_t bigger = strlen(longer), smaller = strlen(shorter); // 3, 2
    if (smaller > bigger) {
        shorter = n1;
        longer = n2;
        size_t temporary = smaller;
        smaller = bigger;
        bigger = temporary;
    }
    *sum = malloc(bigger + 2); // 3 + carry + null byte
    (*sum)[0] = '0';
    (*sum)[bigger + 1] = ''; // ['0', x, x, x, '']
    for (int power_of_ten = bigger; power_of_ten > 0; power_of_ten--) { // 3 ... 1 (for length comparison)
        size_t idx = bigger - power_of_ten; // 0 ... 2 (for indexing left to right)
        if (power_of_ten > smaller) {
            (*sum)[idx + 1] = longer[idx]; // just copy from longer to sum
        } else {
            char c = shorter[idx - (bigger - smaller)] + longer[idx] - '0'; // add
            (*sum)[idx + 1] = c;
            for (int j = idx + 1; j > 0 && (*sum)[j] > '9'; j--) { // backward carry
                (*sum)[j] -= 10;
                (*sum)[j - 1] += 1;
            }
        }
    }
    if ((*sum)[0] == '0') { // ['0', '2', '1', '2'] -> ['2', '1', '2']
        for (int j = 0; j < bigger + 1; j++) {
            (*sum)[j] = (*sum)[j + 1];
        }
    }
    return 42; // problem didn't specify what `int add(...)` returns
}
int main() {
    char a[] = "509843702", b[] = "430958709432";
    char *s;
    (void) add(a, b, &s);
    printf("%sn", s);
    (void) free(s);
    return 1;
}

输出

> ./a.out
431468553134
>

检查

> dc
509843702 430958709432 + p
431468553134
>

此增加涉及五个携带，其中三个是独立的(不是级联。(