关于#c语言#的问题，如何解决？

从键盘输入的字符串由二个大正整数组成的字符串。（123456123456123456 1234567890123456789）。(注意: 整数位数超过12位数，整数之间用空格分开。)
完成：对字符串中的整数进行求和，求积。并且显示计算结果。
用devc++编写

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define MAX_SIZE 1024
typedef struct { char num[MAX_SIZE]; int size; }big_integer_t;
void init_big_integer(big_integer_t* n)
{
    memset(n->num, 0, sizeof(n->num));
    n->size = 0;
}
void add_big_integer(const big_integer_t* lhs, big_integer_t* rhs, big_integer_t* result)
{
    int lhsi = lhs->size - 1, rhsi = rhs->size - 1;
    result->size = 0;
    int addition = 0;
    while (lhsi >= 0 || rhsi >= 0 || addition > 0)
    {
        addition += (lhsi >= 0 ? (lhs->num[lhsi] - '0') : 0)
            + (rhsi >= 0 ? (rhs->num[rhsi] - '0') : 0);
        result->num[result->size++] = (addition % 10) + '0';
        addition /= 10;
        if (lhsi >= 0)lhsi--;
        if (rhsi >= 0)rhsi--;
    }
    result->num[result->size] = 0;
    strrev(result->num);
}

void mul_big_integer(const big_integer_t* lhs, big_integer_t* rhs, big_integer_t* result)
{
    big_integer_t tmp;
    init_big_integer(&tmp);
    result->size = 0;
    for (int i = rhs->size - 1; i >= 0; --i)
    {
        int rn = rhs->num[i] - '0';
        int addition = 0;
        tmp.size = 0;
        for (int j = lhs->size - 1; j >= 0; --j)
        {
            int ln = lhs->num[j] - '0';
            addition += rn * ln;
            tmp.num[tmp.size++] = addition % 10 + '0';
            addition /= 10;
        }
        while (addition > 0)
        {
            tmp.num[tmp.size++] = addition % 10 + '0';
            addition /= 10;
        }
        tmp.num[tmp.size] = 0;

        big_integer_t tmp2 = *result;
        int lhsi = 0, rhsi = (rhs->size - i - 1);
        result->size = rhsi;
        addition = 0;
        while (lhsi < tmp.size || rhsi < tmp2.size || addition > 0)
        {
            addition += (lhsi < tmp.size ? (tmp.num[lhsi] - '0') : 0)
                + (rhsi < tmp2.size ? (tmp2.num[rhsi] - '0') : 0);
            result->num[result->size++] = (addition % 10) + '0';
            addition /= 10;
            if (lhsi < tmp.size)lhsi++;
            if (rhsi < tmp2.size)rhsi++;
        }
    }
    result->num[result->size] = 0;
    strrev(result->num);
}
int add_two_big_integer() {
    char str[2048];
    memset(str, 0, sizeof(str));
    int i = 0, sum = 0, quadrature = 1;
    printf("请输入字符串:\n");
    gets(str);
    int len = strlen(str);
    big_integer_t num0, num1, result_add, result_mul;
    init_big_integer(&num0);
    init_big_integer(&num1);
    init_big_integer(&result_add);
    init_big_integer(&result_mul);
    // 读取第一个大整数
    while (i < len && !isdigit(str[i]))++i;
    while (i < len && isdigit(str[i]))num0.num[num0.size++] = str[i++];
    num0.num[num0.size] = 0;
    // 读取第二个大整数
    while (i < len && !isdigit(str[i]))++i;
    while (i < len && isdigit(str[i]))num1.num[num1.size++] = str[i++];
    num1.num[num1.size] = 0;
    // 相加
    add_big_integer(&num0, &num1, &result_add);
    // 相乘
    mul_big_integer(&num0, &num1, &result_mul);
    printf("字符串中所有正整数之和为：%s,所有正整数之积为：%s\n",
        result_add.num,
        result_mul.num);
    return 0;
}

int main() {
    add_two_big_integer();
}