自己寫的大數相乘

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>


using namespace std;
const int g_kMaxs = 1001;


char g_a[g_kMaxs], g_b[g_kMaxs];


void StrRes(char* str) {
    int len = strlen(str);
    int i = 0, j = len - 1;
    while (i < j) {
        char c = str[i];
        str[i] = str[j];
        str[j] = c;
        ++i;
        --j;
    }
}


int* Mul(char* lhs, char* rhs) {
    if (NULL == lhs || NULL == rhs) { // 爲空則出錯，其他的需要假設保證輸入沒有問題
        fprintf(stderr, "Nul: Invalid arguments");
        exit(1);
    }


    int lhsLen = (int)strlen(lhs);
    int rhsLen = (int)strlen(rhs);
    int* ret = new int[lhsLen + rhsLen + 1];
    // 兩個數相乘之積的位數不會超過兩個數的位數之和，
    // 而申請多一位是把積的長度絕對值放在ret[0]中，並用ret[0]的正負代表積的正負
    memset(ret, 0, sizeof(int)*(lhsLen + rhsLen + 1));
    if ((isdigit(lhs[0]) && isdigit(rhs[0])) || (lhs[0] == '-' && rhs[0] == '-')) {
        ret[0] = 1;
    } else ret[0] = -1;
    if (lhs[0] == '-') --lhsLen;
    if (rhs[0] == '-') --rhsLen;


    StrRes(lhs); //把輸入的數字串反轉
    StrRes(rhs);


    for (int i = 0; i < lhsLen; i++) {
        for (int j = 0; j < rhsLen; j++) {
            ret[i + j + 1] += ((lhs[i] - '0') * (rhs[j] - '0')) % 10;
            ret[i + j + 2] += ((lhs[i] - '0') * (rhs[j] - '0')) / 10;
        }
    }
    StrRes(lhs); //把輸入的數字串反轉恢復
    StrRes(rhs);
    if (ret[0] < 0) printf("-"); // ret[0] < 0 爲負數
    ret[0] = ret[0] * ((ret[lhsLen + rhsLen] > 0) ? (lhsLen + rhsLen) : (lhsLen + rhsLen - 1));
    for (int i = abs(ret[0]); i >= 1; i--) printf("%d", ret[i]);
    printf("\n");
    return ret;
}


int main()
{
    while (EOF != scanf("%s%s", g_a, g_b)) {
        Mul(g_a, g_b);
    }
    return 0;
}