华为机试题：超长正整数相加

描述:  
题目描述：请设计一个算法完成两个超长正整数的加法。  
要求实现函数： 
void AddLongInteger(char * pcAddend, char * pcAugend, char * pcAddResult); 
输入参数： 
char * pcAddend：加数 
char * pcAugend：被加数 
char * pcAddResult：加法结果 
返回值：无 
运行时间限制: 1 Sec 
内存限制: 128 MByte 
输入:  
两个超长正整数的字符串 
输出:  
相加后结果的字符串 
样例输入: 
123456789123456789 123456789123456789 
样例输出: 
246913578246913578

测试不通过，发出来看看问题在哪里。

//
//  超长正整数相加.cpp
//  temptest
//
//  Created by 彭威 on 15/8/1.
//  Copyright © 2015年 biophy.nju.edu.cn. All rights reserved.
//

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;

class BigNumber {
    string num;
    bool reverse_flg;

public:
    BigNumber(char *str):num(str){}
    BigNumber(){};
    friend istream& operator>>(istream& in,BigNumber& p){
        in>>p.num;
        return in;
    }
    friend ostream& operator<<(ostream& out,const BigNumber& p){
        if (p.reverse_flg) {
            //for (auto iter = p.num.rbegin(); iter != p.num.rend(); iter++) {       //only for c++11
            for (string::const_iterator iter = p.num.end()-1;iter >= p.num.begin();--iter ){
                out<<*iter;
            }
        }else{
            out<<p.num;
        }
        return out;
    }
    friend const BigNumber operator+(const BigNumber& p1,const BigNumber& p2){
        BigNumber p;
        size_t len = max(p1.length(), p2.length());
        char plus=0;
        p.reverse_flg = true;
        for (size_t i =0;  i < len; i++) {
            char x = p1[i] + p2[i] + plus - 2 * ‘0‘;
            plus = x / 10;
            x = x % 10 + ‘0‘;
            p.num.push_back(x);
        }
        if (plus) {
            p.num.push_back(plus+‘0‘);
            plus = 0;
        }
        return p;
    }
    const size_t length() const{
        return num.length();
    }
    const char operator[](size_t n)const {
        if (n >= num.length()) {
            return ‘0‘;
        }
        if (reverse_flg) {
            return num[n];
        }else{
            return num[num.length()-n-1];
        }
    }
};

int main(int argc,const char* argv[]){
    BigNumber a,b;
    cin>>a>>b;
    cout<<a+b<<endl;
    return 0;
}