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

Answer is to use binary AND.

so a number is represented in memory in 0 and 1. lets say 4 and 5.

4 = 0000 0100

5 = 0000 0101

and every even number has a zero in the end and every odd number has 1 in the end;

in c '1' means true and '0' means false.

so: lets code;

function isEven(int num){
     return ((num & 0x01) == 0) ? 1 : 0;
}

Here 0x01 means 0000 0001. so we are anding 0x01 with the given number.

imagine no is 5

5    |0000 0101 

0x01 |0000 0001

---------------

      0000 0001

so answer will be '1'.

imagine no is 4

4    |0000 0100 

0x01 |0000 0001

---------------

      0000 0000

so answer will be '0'.

now,

return ((num & 0x01) == 0) ? 1 : 0;

it is expanded in :

if((num & 0x01) == 0){// means  the number is even
      return 1;
}else{//means no is odd
      return 0;
}

So this is all.

End is binary operatiors are very important in compititive programming world.

happy coding.

first answer here.

EDIT 1:

Total no of evens are

totalEvens = ((end - start) / 2 + ((((end - start) & 0x01 ) == 0) ? 0 : 1 ));

here (end - start)/2 gives the half of total numbers.

this works if one is even and one is odd.

but,

((((end - start) & 0x01 ) == 0) ? 0 : 1 )

can just replaced by (!isEven(end-start))

So, if the total number is even then dont add 1 else add 1.

this completely works.