The following code seems to overflow. Why?
#include <chrono>
#include <iostream>
int main() {
auto now = std::chrono::steady_clock::now();
auto low = std::chrono::steady_clock::time_point::min();
auto elapsed = std::chrono::duration_cast<std::chrono::seconds>(now - low).count();
std::cout << elapsed << "\n";
return 0;
}
I would expect low to be the earliest time point. That's how I understand the "Return value" description under https://en.cppreference.com/w/cpp/chrono/time_point/min.
So, any other value, e.g. now, should be greater and the difference now - low should be a positive duration. Unless it somehow overflows somewhere? Is it a bug (unlikely; all main compilers seem to have this issue)? Is it my misunderstanding of chrono somehwere that is at fault?
How can I fix this?
Context: I have a list of elements. Some elements are actively being used, others have a timestamp which tells when they were last used. Periodically, timestamps are checked if they are too old, and if so elements with those early timestamps are removed from the list.
A user can manually invoke a flush operation. It sets all timestamps that exist to low so that when the next check is performed, all those elements are removed.
I can set low to be now - reasonableBigButNotTooBig value to make it work, but I thought std::chrono::steady_clock::time_point::min would be more idiomatic.
std::chrono::steady_clock::time_point::min() isn't generally zero, usually std::chrono clocks use a signed 64-bit integer as the representation of the time_point. low is therefore likely to be -2^63 ticks, now is likely a positive number of ticks therefore now - low, or equivalently now + 2^63, is likely to overflow.
The epoch of steady_clock isn't specified but in general all timestamps will be positive (the epoch is generally the boot time of the computer) so you can use std::chrono::steady_clock::time_point{} as a "minimum" time point instead.