Time duration Calculator with c++ Problems - c++

Question

I'm currently taking a c++ college course, and I'm running into a problem that makes me feel like an idiot. We're making a time duration calculator that calculates the time after and before a certain duration, and the assignment tries to instruct us how to avoid negatives. I'm not sure what I'm doing wrong, but no matter what I do, it won't work. This is his instruction: "Also, when negative operands are used, the remainder operator is not defined the same way that the mathematical modulus operation (Euclidian Division) is defined. For those reasons, it is easier to just make sure we don't end up with negative remainders. This is easily done by adding a day (or two or three) to before when calculating the difference. Go ahead and do that."
and this is my code (all of which has followed his instructions till now)
#include <iostream>
#include <string>
#include <iomanip>
using namespace std;
int run()
{
int timeHours;
int timeMinutes;
int durationHours;
int durationMinutes;
char discard; // won't keep this character
cout << "Enter current time: " << endl;
cin >> timeHours >> discard >> timeMinutes;
cout << "Enter duration: " << endl;
cin >> durationHours >> discard >> durationMinutes;
// processing section - compute the results
int timeInMin = (timeHours * 60) + timeMinutes;
int durationInMin = (durationHours * 60) + durationMinutes;
int after = timeInMin + durationInMin;
int before = timeInMin - durationInMin;
int afterHours = after / 60%12;
int afterMinutes = after % 60;
int beforeHours = (before / 60);
int beforeMinutes = (before % 60);
// output section: test data inside brackets[]
cout << setfill('0'); // only needed once
cout << endl;
cout << durationHours << ":" << setw(2) << durationMinutes << " hours after, and before, " <<
timeHours << ":" << setw(2) << timeMinutes << " is [" << afterHours << ":" << setw(2) <<
afterMinutes << ", " << beforeHours << ":" << setw(2) << beforeMinutes << "]" << endl;
return 0;
}
for example, if I plug in 2:30 as the time and 13:27 as the duration, expected [3:57, 1:03] but found [3:57, -10:-57]
I've tried adding both 24 and 1440 (number of minutes in a day) to before as well as the calculations that use the int before, but no luck. I've tried reaching out to my class a few days ago, but no one has responded.
Edit: he also says to do this in order to get rid of midnight times that look like 0:32 by doing this: Add 11 to after-hours. Find the remainder of dividing by 12. Add 1 to the result. Does he mean something like:
int afterHours = (((after / 60%12) + 11) / 12) + 1
the %12 was so that the hours wouldn't go over 12 and turn into military time

Answer 1

Instead of this:
// processing section - compute the results
int timeInMin = (timeHours * 60) + timeMinutes;
int durationInMin = (durationHours * 60) + durationMinutes;
int after = timeInMin + durationInMin;
int before = timeInMin - durationInMin;
int afterHours = after / 60%12;
int afterMinutes = after % 60;
int beforeHours = (before / 60);
int beforeMinutes = (before % 60);
This:
int timeInMin = (timeHours * 60) + timeMinutes;
int durationInMin = (durationHours * 60) + durationMinutes;
int after = timeInMin + durationInMin;
int before = (timeInMin % 720) + 720 - durationInMin; // 720 is 12 hours in minutes
int afterHours = (after / 60) % 12;
int afterMinutes = (after - (afterHours * 60)) % 60;
afterHours = afterHours + 12 * (afterHours == 0);
int beforeHours = (before / 60) % 12;
int beforeMinutes = (before - (beforeHours * 60)) % 60;
beforeHours = beforeHours + 12 * (beforeHours == 0);

Answer 2

The trick is to get an integer into the range [1,12] using only +, - and %.
Let's start with the negative numbers:
t = 8;
d = 11;
b = t-d; // -3
So we add 12:
b = t-d + 12; // 9
Ah, but what if the interval is bigger?
d = 100;
b = t-d + 12; // -80
This is why your teacher talked about adding "a day or two or three". But how to know how many? No need, we have modulo:
d = 100;
d = d%12; // 4
b = t-d + 12; // 16
Now we're positive, but we're getting numbers greater than 12. No problem, we use modulo again:
d = 100;
d = d%12; // 4
b = (t-d + 12)%12; // 4
a = (t+d)%12; // 0
Ah, now we're getting [0,11], but we want [1,12]. So we use the trick (x-1)%m +1:
d = 100;
d = d%12; // 4
b = (t-d + 12 - 1)%12 + 1; // 4
a = (t+d - 1)%12 + 1; // 12
There we have it. We still have to deal with minutes (the same way) and with a smart Alec user who enters negative numbers, but it should be clear how to do that and this post is getting long.