Re: [OT]Re: rounding to a multiple of x?
Re: [OT]Re: rounding to a multiple of x?
- Subject: Re: [OT]Re: rounding to a multiple of x?
- From: "Edward K. Chew" <email@hidden>
- Date: Wed, 19 May 2004 14:58:41 -0400
On May 19, 2004, at 13:07, Graham J Lee wrote:
On 2004-05-19 16:47:23 +0100 email@hidden wrote:
Greetings,
I can't seem to figure out how to roundup to a multiple of a number
ie if i have 1149 and i want to round up to a multiple of 4 how would
accomplish this? any direction would be great
If we're talking integers, and we call 1149 'x' and 4 'y'...
y*(x/y+1)
But if x and y were both 4, you'd be unnecessarily rounding up to 8,
right? How about this:
((x + y - 1) / y) * y
I think it'll work as long as x >= 0. For negative numbers, the y - 1
would have to go, so the general case for integers might look like
this:
((x < 0 ? x : x + y - 1) / y) * y
Hmm...I don't particularly like that conditional logic in there,
though. Maybe there's a better way. If you knew that y was always a
power of 2, you could use some bit logic like so:
(x + y - 1) & -y
That should even work for negative x values, I think.
-Ted
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