C++11 std::chrono subtract now and min

Question

I feel like I'm going a little crazy with this one, but it just doesn't make sense to me. In my mind, if I subtract the minimum time point from any time point returned from a now() call, I should always get a positive duration, but that doesn't happen.

#include <chrono>
#include <iostream>

typedef std::chrono::steady_clock myclock;

int main(int argc, char **argv) {
        myclock::time_point min = myclock::time_point::min();
        myclock::time_point now = myclock::now();
        auto millis = std::chrono::duration_cast<std::chrono::milliseconds>(now - min).count();
        std::cout << millis << std::endl;
}

Why does this print a negative integer and not a positive integer? (clang 3.3 or g++ 4.8.1)

Answer 1

As has been pointed out this is the result of overflow. Remember that the minimum value a signed type can represent is about the same magnitude as the largest. If now is positive then the difference between now and min will clearly have greater magnitude than min, which means it has greater magnitude than the type's largest value can represent.

If you want to guarantee a positive duration then instead of using the minimum you could instead use a steady clock and then use the program start time as the base. The built-in clock durations are all specified such that a duration should be able to represent a range of at least a couple hundred years, so unless your program runs for longer than that you'll get positive durations.

Another option is to choose a clock where the epoch is known to be in the past and simply say

Clock::now().time_since_epoch();

Answer 2

You're overflowing the counter which, on my machine, is a signed long long

#include <chrono>
#include <iostream>
#include <limits.h>
using namespace std;

typedef std::chrono::steady_clock myclock;

int main(int argc, char **argv) {
    myclock::time_point min = myclock::time_point::min();
    long long minl = reinterpret_cast<long long&>(min);
    cout << reinterpret_cast<long long&>(min) << endl;

    auto now = myclock::now();
    long long nowl = reinterpret_cast<long long&>(now);
    cout << reinterpret_cast<long long&>(now) << endl;

    cout << (nowl-minl) << endl;

    cout << "max of signed long long:" << LLONG_MAX << endl;

    auto millis = std::chrono::duration_cast<std::chrono::seconds>(now - min).count();
    //std::cout << millis << std::endl;
}

Output:

-9223372036854775808
13924525439448122
-9209447511415327686
max of signed long long:9223372036854775807