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

Oh well, why not:

#include <cassert>

int ecount( int begin, int end ) {
    assert( begin <= end );
    int size = (end - begin) + 1;
    if ( size % 2 == 0  || begin % 2 == 1 ) {
        return size / 2;
    }
    else {
        return size / 2 + 1;
    }
}

int main() {
    assert( ecount( 1, 5 ) == 2 );
    assert( ecount( 1, 6 ) == 3 );
    assert( ecount( 2, 6 ) == 3 );
    assert( ecount( 1, 1 ) == 0 );
    assert( ecount( 2, 2 ) == 1 );
}

Answer 2

int start, stop;
start = 0;
stop = 9;
printf("%d",(stop-start)/2+((!(start%2) || !(stop%2)) ? 1 : 0));

Where start and stop can hold any value. No need to iterate to to determine this number.

Answer 3

This'll do the trick, even for ranges with negative numbers.

int even = (last - first + 2 - Math.abs(first % 2) - Math.abs(last % 2)) / 2;

Tested with the following code:

public static void main(String[] args) {
    int[][] numbers = {{0, 4}, {0, 5}, {1, 4}, {1, 5}, {4, 4}, {5, 5},
                       {-1, 0}, {-5, 0}, {-4, 5}, {-5, 5}, {-4, -4}, {-5, -5}};

    for (int[] pair : numbers) {
        int first = pair[0];
        int last = pair[1];
        int even = (last - first + 2 - Math.abs(first % 2) - Math.abs(last % 2)) / 2;
        System.out.println("[" + first + ", " + last + "] -> " + even);
    }
}

Output:

[0, 4] -> 3
[0, 5] -> 3
[1, 4] -> 2
[1, 5] -> 2
[4, 4] -> 1
[5, 5] -> 0
[-1, 0] -> 1
[-5, 0] -> 3
[-4, 5] -> 5
[-5, 5] -> 5
[-4, -4] -> 1
[-5, -5] -> 0

Answer 4

This does not require any conditions at all:

evencount = floor((max - min)/2) + 1

Answer 5

I'd say

(max - min + 1 + (min % 2)) / 2

Edit: Erm okay for some reason I thought that (min % 2) returns 1 for even numbers.... :). The correct version is

(max - min + 1 + 1 - (min % 2)) / 2

or rather

(max - min + 2 - (min % 2)) / 2

Answer 6

The first even number in the range is: (begin + 1) & ~1 (round begin up to even number).

The last even number in the range is: end & ~1 (round end down to even number).

The total number of even numbers in the range is therefore: (end & ~1) - ((begin + 1) & ~1) + 1.

int num_evens = (end & ~1) - ((begin + 1) & ~1) + 1;