c++ - 如何将两个 'hours' 作为整数相加/相减？

Question

我知道这可能是个愚蠢的问题，但我遇到了一些麻烦，我很惭愧，但我真的不知道如何做到。我想加减两个以整数形式给出的“小时”。

#include <iostream>
#include <cstdlib>
#include <cmath>
using namespace std;

int main()
{
    // should be: 21:59:2
    int rH, rM, rS;

    int h1 = 12, m1 = 34, s1 = 56;
    int h2 = 9, m2 = 24, s2 = 6;

    if(h1 + h2 >= 24)
    {
        rH = abs(h2 + h1);
        m1 += 60;

        if(m1 + m2 >= 60)
        {
            rH = (m2 - m1);
        }

    }
    else
    {
        rH = h1 + h2;
    }

    cout << rH << " " << rM << " " << rS << " " << "\n";

    return 0;
}

试过上面的但我不知道如何进一步做。我不能使用任何 date-time 对象。

编辑

减法，我不知道错误在哪里但计算错误:

#include <iostream>
#include <cstdlib>
#include <cmath>
using namespace std;

int main()
{
    int rH, rM, rS;
    int h1 = 12, m1 = 25, s1 = 29, h2 = 11, m2 = 12, s2 = 1;
    int sum1, sum2, sum;

    sum1 = (h1)*3600 + (m1)*60 + s1;
    sum2 = (h2)*3600 + (m2)*60 + s2;

    if(sum1 > sum2)
        sum = sum1-sum2;
    else
        sum = sum2-sum1;

    rS = sum %60;
    sum = sum/60;
    rM = sum % 60;
    rH = sum/60;

    std::cout << rH << " " << rM << " " << rS << " " << "\n";

    return 0;
}

Answer 1

为什么不

int t1 = ((h1 * 60) + m1) * 60 + s1;
int t2 = ((h2 * 60) + m2) * 60 + s2;
int tdiff = t1 - t2;
if (tdiff < 0) { tdiff += 24 * 60 * 60;

然后使用模块化算法和除法来获取小时数等，例如

int tdiffs = tdiff % 60;
tdiff  = tdiff / 60;
int tdiffm = tdiff % 60;
tdiff h = tdiff / 60;