Smallest integer multiple of a number >= another number

[sethmachine]

Hi,

Here is an example:

a = 4

b = 7

foo(4, 7) = 2, because 2 is the smallest multiple of 4 which is greater than or equal to 7 (2 * 4 = 8 >= 7)

Another example:

a = 2

b = 5

foo(2, 5) = 3, because 3 is the smallest multiple of 2 which is greater than or equal to 5 (3 * 2 = 6 >= 5)

How to get this function in JASS?

 

[Ceday]

function foo takes integer a, integer b returns integer
    /* 
    error check if you want
    if a <= 0 or b <= 0 then
        return 0
    elseif a > b then
        return foo(b, a)
    endif
    */

    if (b/a * a == b) then
        return b/a
    else
        return b/a + 1
    endif
endfunction

 

[muzzel]

Ur looking for "ceil(b/a)" http://en.wikipedia.org/wiki/Floor_and_ceiling_functions

function foo takes integer a, integer b returns integer
    return R2I(1. * b / a + 0.5)
endfunction

 

[sethmachine]

Thanks Ceday and muzzel.

I guess I just forgot simple math.

 

[muzzel]

Actually "(b+a-1)/a" should be faster, as R2I is really slow. Not that it matters...

function foo takes integer a, integer b, returns integer
    return (b + a - 1) / a
endfunction

 

[Dr Super Good]

(b + a - 1) / a

This may return nonsense if both b and a are very large and approaching the limits of int32_t integer type that JASS integer are.

a = 2147473647
b = 2147478647
Answer should be 2
(2147478647 + 2147473647 - 1) = -15003 // Oh dear we have overflowed...
-15003 / 2147473647 = 0 // not the right answer.

 

[sethmachine]

I doubt a, b will ever be higher than 100 (I don't see them even reaching ~20). So it shouldn't be an issue, but thank you for pointing this out as a note when using muzzel's solution.