如何四舍五入到另一个数字的最接近的分数？

Question

I want to add a special scaling mode where it snaps to the closest multiple. Right now, scaling up is fine, it's linear and goes something like this:

 // f(x := 35, n := 32) = 32
 // f(x := 66, n := 32) = 64
 x = floor(x / n) * n

But for scaling down, I want some sort of inverse exponential:

 // f(x := 18, n := 32) = 16
 // f(x := 12, n := 32) = 8
 // f(x := 7, n := 32) = 4

An implementation suggestion without using a loop in any language would suffice. My current implementation is as follows (which 100% works):

float ScaleMultiple(float value, float multiple) {
    if (value > multiple) {
        return(floor(value / multiple) * multiple);
    } else if (value < multiple) {
        do {
            multiple /= 2.0f;
        } while (value >= multiple);
        return(multiple);
    } else {
        return(value);
    }
}

multiple is a positive floating point number and not necessarily a multiple of 2 and is at least equal to 1. value is a positive floating point number and is at least equal to 1.

Answer 1

32 / 18 ~= 1.8 ~= 2, the result you want is 32 / 2, right? For the case of 12 and 32, this would yield 3 though, not four. In order to get there, take the log-2 of that number, round that to an even number and then convert back. However, this has the problem that it doesn't work properly when n is not a power of two.

Answer 2

You can get the correct results for the values you give in Java as follows:

int f = (int)Math.pow(2,(int)(Math.log(x)/Math.log(2)));

There's insufficient information provided to answer the question for arbitrary n.