Question

r, a and b are integers.

I need the cheapest computation since in a critical part of the code.
I found :

r = (a / b) + (((a % b) != 0) ? 1 : 0);

if b is a power of 2, then a / b can be replaced with a >> log2(b)

and a % b with a & (b-1) which should save a lot of computation time.

Do you know any better solution ?

Answer 1

val r = (a + b - 1) / b

For example:

scala> for(a <- 1 to 10; b <- 1 to a) println("a: "+a+"\tb: "+b+"\tr: "+((a+b-1)/b))
a: 1    b: 1    r: 1
a: 2    b: 1    r: 2
a: 2    b: 2    r: 1
a: 3    b: 1    r: 3
a: 3    b: 2    r: 2
a: 3    b: 3    r: 1
a: 4    b: 1    r: 4
a: 4    b: 2    r: 2
a: 4    b: 3    r: 2
a: 4    b: 4    r: 1
a: 5    b: 1    r: 5
a: 5    b: 2    r: 3
a: 5    b: 3    r: 2
a: 5    b: 4    r: 2
a: 5    b: 5    r: 1
a: 6    b: 1    r: 6
a: 6    b: 2    r: 3
a: 6    b: 3    r: 2
a: 6    b: 4    r: 2
a: 6    b: 5    r: 2
a: 6    b: 6    r: 1
a: 7    b: 1    r: 7
a: 7    b: 2    r: 4
a: 7    b: 3    r: 3
a: 7    b: 4    r: 2
a: 7    b: 5    r: 2
a: 7    b: 6    r: 2
a: 7    b: 7    r: 1
a: 8    b: 1    r: 8
a: 8    b: 2    r: 4
a: 8    b: 3    r: 3
a: 8    b: 4    r: 2
a: 8    b: 5    r: 2
a: 8    b: 6    r: 2
a: 8    b: 7    r: 2
a: 8    b: 8    r: 1
a: 9    b: 1    r: 9
a: 9    b: 2    r: 5
a: 9    b: 3    r: 3
a: 9    b: 4    r: 3
a: 9    b: 5    r: 2
a: 9    b: 6    r: 2
a: 9    b: 7    r: 2
a: 9    b: 8    r: 2
a: 9    b: 9    r: 1
a: 10   b: 1    r: 10
a: 10   b: 2    r: 5
a: 10   b: 3    r: 4
a: 10   b: 4    r: 3
a: 10   b: 5    r: 2
a: 10   b: 6    r: 2
a: 10   b: 7    r: 2
a: 10   b: 8    r: 2
a: 10   b: 9    r: 2
a: 10   b: 10   r: 1

This does assume a and b are positive. If either are negative, it depends on whether the division is symmetric or floored (modern languages and platforms are symmetric), and the signal of a and b.

If a*b >= 0, then the formula works as given. If the division is symmetric and a*b < 0, then a / b gives the correct answer.