c - 如何对两个包含长数字的字符串进行加法？在C中

Question

char* a = "5880391469248794735212"
char* b = "1234567890231"

我需要进行加法计算并将新数字放入字符串中。
我尝试将每个数字转换为整数，将它们相加并将结果转换为字符串，但数字的最大值是 unsigned long long ，它不能包含我显示的数字。

感谢您的帮助!

Answer 1

您应该使用像 GMP 这样的大数字库，但如果您愿意，也可以尝试以下代码:

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

char *add(char *x, char *y);

int main(void)
{
    char *a = "5880391469248794735212";
    char *b = "1234567890231";

    char *c = add(a, b);

    printf("z: %s\n", c);

    free(c);
    return 0;
}

void reverseString(char *buf)
{
    char *ptr = buf + strlen(buf) - 1;

    while (buf < ptr) {
        char tmp = *buf;
        *buf++ = *ptr;
        *ptr-- = tmp;
    }
}

char *add(char *a, char *b)
{
    size_t xlen = strlen(a);
    size_t ylen = strlen(b);

    size_t n = 0;

    int val;
    int r = -1;

    char *p1, *p2;
    char *t1, *t2;

    char *c = NULL;

    if (xlen >= ylen) {
        p1 = a + xlen - 1;  t1 = a;
        p2 = b + ylen - 1;  t2 = b;
        c = (char *)malloc(xlen + 2);
    }
    else {
        p1 = b + ylen - 1;  t1 = b;
        p2 = a + xlen - 1;  t2 = a;
        c = (char *)malloc(ylen + 2);
    }
    if (!c) {
        fputs("error: memory allocation failed.\n", stderr);
        return NULL;
    }

    while (p2 >= t2)
    {
        if (r == -1) {
            val = (*p1-- - '0') + (*p2-- - '0');
        }
        else {
            val = (*p1-- - '0') + (*p2-- - '0') + 1;
        }
        if (val > 9) {
            r = val - 10;
            c[n++] = r + '0';
        }
        else {
            c[n++] = val + '0';
            r = -1;
        }
    }

    while (p1 >= t1)
    {
        if (r == -1) {
            c[n++] = *p1--;
        }
        else {
            val = (*p1-- - '0') + 1;
            if (val > 9) {
                r = val - 10;
                c[n++] = r + '0';
            }
            else {
                c[n++] = val + '0';
                r = -1;
            }
        }
    }
    if (r != -1) {
        c[n++] = '1';
    }
    c[n] = '\0';
    reverseString(c);
    return c;
}