evaluate an odd number

[winecoding]

From a technical blog, I learned a trick to check whether i is odd number by evaluating

(i & 1) != 0

I am not sure how it works?

[m4ster r0shi]

(i & 1) evaluates to 1 if the least significant bit of i is 1, and to 0 otherwise. The thing here is that a number is odd if-f its least significant bit is 1. Think about it...

A binary number looks like this:

... d_n ... d₂ d₁ d₀, where d_k are the individual digits, each being either one or zero.

The value of this number is:

... + 2ⁿ * d_n + ... + 2² * d₂ + 2¹ * d₁ + 2⁰ * d₀

As you can see, in the above sum, every term other than 2⁰ * d₀ is always even.
Thus, if d₀ is 0 the number is even and if d₀ is 1 the number is odd.

[Galik]

If you look at numbers in binary form you will notice that the least significant column alternates between 0 and 1. So all even numbers end in 0 and all odd numbers end in 1 in binary.


0000 = 0
0001 = 1
0010 = 2
0011 = 3
0100 = 4
0101 = 5
0110 = 6
0111 = 7
1000 = 8
1001 = 9
1010 = 10
1011 = 11
1100 = 12
1101 = 13
1110 = 14
1111 = 15



So the simplest wa to check if a number is odd or even is by examining the least significant bit. This is done using the bitwise AND operator &

[winecoding]

Thanks for your explanations.