how to avoid the potential of an overflow when computing an average of times?

Question

I am writing a function for getting the average of the clocks it takes to call a specific void (*)(void) aka void -> void function a specific number

Answer 1

The average of the first n elements is

          SUM
Average = ---
           n

The next element Mi is

           (SUM + Mi)
Average2 = ----------
              n + 1

So given the current average, it is possible to find the next average with the new reading.

           (Average * n + Mi )
Average2 = -------------------
                  n + 1

This can then be changed to an equation which doesn't increase

                       n      Mi
Average2 = Average * ----- + -----
                     n + 1   n + 1

In practice for timing, the size of time will fit within the datatype of the computer.

As pointed out, this needs to use a floating point representation, and whilst will not fail due to overflow, can still fail when n/(n+1) is smaller than the accuracy of the floating point fraction part.

Update

From incremental average

There is a better reorganization.

                       Mi - Average
Average2 = Average  +  -------------
                           n + 1

It is better, because it only has one division.