timer using chrono in C++

The default datatypes used by std::chrono::duration and std::chrono::timepoint are integers (unspecified which one but usually int64t).

Adding std::chrono::duration to std::chrono::systemclock::timepoint will result in a std::chrono::timepoint.

The iostream output operator for systemclock timepoints is explicitly disabled for floating point durations which is why you can't print it.

std::chrono::systemclock::timepoint end = start + std::chrono::durationcast(timer_duration);

It has been rightly pointed out by Alan Birtles and Comet, the duration value needs to be an integer type to enable straightforward addition of 'start' and 'duration' and subsequent printing. However, I wanted to allow the user to input timer duration as a fraction of a second. So I save the double input into nanoseconds and save that as an integer duration. Changes made:

std::chrono::nanoseconds timerduration((long long)(timer*1000000000));

std::chrono::timepoint  start, end;
//...
end = start + timer_duration;
std::cout 

Question 1 :
You declare start as a std::chrono::timepoint but end is initialised as a addition of start and timerduration which is declared as auto. Auto deduced type with initialization: here start is an int and timeduration a double so the sum is of type double and the auto deduction creates a time point (std::chrono::timepoint) in double. Check (What is the meaning of the auto keyword?) for more explanation.

That's what explains the difference of type between start and end. So, in conclusion, if you want start and end to be of the same type and calculate duration:

std::chrono::timepoint start, end;
std::chrono::duration timerduration(timer); 
start = std::chrono::systemclock::now();
end = start + std::chrono::durationcast(timer_duration);

Question 2 :
Report to question 1, end was a double see more detailed answer of Alan Birtles.