Simplest way to calculate amount of even numbers in given range

Question

What is the simplest way to calculate the amount of even numbers in a range of unsigned integers?

An example: if range is [0...4] then the answer is 3 (0,2,4)

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Answer 1

The count of even numbers between 0 and n is [n/2] + 1. Therefore the count of even numbers between (n + 1) and m is ([m/2] + 1) - ([n/2] + 1) = [m/2] - [n/2].

For count of even numbers between m and n the answer therefore would be [m/2] - [(n - 1)/2].

The [x] is taken to the direction of -\infty. Beware that the usual C integer division is not doing right in our case: a/2 is rounded towards zero, not -\infty, so the result will be not [a/2] for teh case of negative a.

This should be the simplest calculation; works for negative numbers, too. (Needs however that m >= n.) Doesn't contain ifs and ?:s.

If you don't consider negative numbers, you can use just m/2 - (n+1)/2 + 1, otherwise floor(m/2.0) - floor((n-1)/2.0)