CODEWITHSUNDEEP
c++c++-chrono
Andrew
Andrew

Reputation: 658

Period count function

I'm new to the chrono library and I'm trying to use it to write a function that returns how many 5 min periods there are between two times. Please note: The 1min source and 5min result period could be different so the solution needs to be flexible (e.g. 5min source, 15min result period is one possible combination).

For example, I have the following general code:

using namespace std::chrono;
using namespace std::chrono_literals;

auto tp_now = system_clock::now();  //define 1st tp
auto tp_test = tp_now - 4min;       //define 2nd tp

duration<int64_t, std::ratio<1, 5*60>> five_mins;   //define a duration of 5mins
std::cout << "period_index = " << floor(five_mins(tp_test)).count() << std::endl;  //print out the 
                                                                                   //period 
                                                                                   //duration count

This code does not compile (I'm not sure why) but it hopes to display how many 5 min periods are within the 4min test period. Clearly, the 4min test period is a variable and could be different. Some examples of where the 4min variable is different with corresponding results are given below:

0-4 mins    result = 0
5-9 mins    result = 1
10-14 mins  result = 2
15-19 mins  result = 3
20-24 mins  result = 4

How can I achieve this please?

Upvotes: 1

Views: 190

Answers (2)

user4442671
user4442671

Reputation:

five_mins should be a type, not a variable.

Also, std::ratio<1, 5*60> makes for 1 three-hundredths of a second, not 5 minutes.

using five_mins = duration<int64_t, std::ratio<5*60>>;

Furthermore, since there is potential truncation happening and the time is not in floating point, you cannot use the five_mins constructor. You HAVE to use std::chrono::duration_cast<>, like so:

#include <chrono>
#include <iostream>

int main() {
    using five_mins = std::chrono::duration<int64_t, std::ratio<5*60>>;
    std::cout << std::chrono::duration_cast<five_mins>(std::chrono::seconds(700)).count() << std::endl;
}

You can see the formal details here: https://en.cppreference.com/w/cpp/chrono/duration/duration, under constructor #4.

Finally, from your original question, you still need to convert your time point into a duration to get a fully working program:

std::cout << "period_index = " << std::chrono::duration_cast<five_mins>(tp_test-tp_now).count() << std::endl

Upvotes: 6

Marshall Clow
Marshall Clow

Reputation: 16680

is there some reason you don't want to write:

std::cout << (/*some duration*/ / minutes(5)) << std::end;

See CppReference

For example, the value of minutes(14)/minutes(5) is 2, and the value of seconds(1000) / minutes(5) is 3

Upvotes: 3

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