大数加法C++实现

题目：假设输入是2个数字，可能超过long long类型能表示的范围，请输出两数相加的运算结果。

思路：2个数输入的时候，肯定都是用string存的，先将短的数在末尾补0，使得二者一样长。然后挨个位相加，并设置一个标志位专门用来存溢出(即进位)情况。
代码如下：

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

using namespace std;

static void compasent0(std::string& inputStr1, std::string& inputStr2)
{
    if (inputStr1.size() < inputStr2.size())
    {
        int diff = inputStr2.size() - inputStr1.size();
        string prefix(diff, '0');
        inputStr1 = prefix + inputStr1;
    }
    else
    {
    {
        int diff = inputStr1.size() - inputStr2.size();
        string prefix(diff, '0');
        inputStr2 = prefix + inputStr2;
    }
    }
}

static bool checkStrValid(const std::string inputStr)
{
    for(const char& ch : inputStr)
    {
        if ((ch < '0') || (ch > '9'))
        {
            return false;
        }
    }
    return true;
}

static std::string BigNumAdd(std::string inputStr1, std::string inputStr2)
{
    if (!checkStrValid(inputStr1) || !checkStrValid(inputStr2))
    {
        printf("ERROR, invalid input!\n");
        return "ERROR";
    }

    compasent0(inputStr1, inputStr2);
    printf("line[%u], inputStr1[%s], inputStr2[%s]\n", __LINE__, inputStr1.c_str(), inputStr2.c_str());

    std::string out(inputStr1.size(), '0');
    int tempSum = 0;
    int overflow = 0;
    for (int i = inputStr1.size() - 1; i > -1; --i)
    {
         tempSum = (int)(inputStr1[i] - '0') + (int)(inputStr2[i] - '0') + overflow;
         if (tempSum < 10)
         {
            out[i] = (char)(tempSum + '0');
            overflow = 0;
         }
         else
         {
            out[i] = (char)(tempSum % 10 + '0');
            overflow = 1;
         }
    }
    if (overflow)
    {
        out = to_string(overflow) + out;
    }

    return out;
}

int main()
{
    std::string inputStr1 = "1906";
    std::string inputStr2 = "456";
    std::string outputStr = BigNumAdd(inputStr1, inputStr2);
    printf("outputStr[%s]\n", outputStr.c_str());

    inputStr1 = "99999999999999999999";
    inputStr2 = "1";
    outputStr = BigNumAdd(inputStr1, inputStr2);
    printf("outputStr[%s]\n", outputStr.c_str());

    return 0;
}

输出是：

./a.out
line[45], inputStr1[1906], inputStr2[0456]
outputStr[2362]
line[45], inputStr1[99999999999999999999], inputStr2[00000000000000000001]
outputStr[100000000000000000000]