Question

To find the next odd number for an input the following code is being used:

a=5.4; // Input
b=Math.ceil(a); // Required to turn input to whole number 
b=b+(((b % 2)-1)*-1); // Gives 7

The ceil rounding function is required.

Is this safe and is there a more compact way to do this?

EDIT: When the input is already an odd whole number then nothing happens. For example 5.0 will return 5

Answer 1

At the question author’s request:

The most compact way to achieve it is

b = Math.ceil(a) | 1;

First use ceil() to obtain the smallest integer not smaller than a, then obtain the smallest odd integer not smaller than ceil(a) by doing a bitwise or with 1 to ensure the last bit is set without changing anything else.

To obtain the smallest odd integer strictly larger than a, use

b = Math.floor(a+1) | 1;

Caveats:

Bit-operators operate on signed 32-bit integers in Javascript, so the value of a must be smaller than or equal to 2^31-1, resp. strictly smaller for the second. Also, a must be larger than -2^31-1.

If the representation of signed integers is not two’s complement, but ones’ complement or sign-and-magnitude (I don’t know whether Javascript allows that, Java doesn’t, but it’s a possibility in C), the value of a must be larger than -1 — the result of Math.ceil(a) resp. Math.floor(a+1) must be nonnegative.