在Java中选择随机颜色

Question

I have an iPhone version of my application which is using a bit of code for setting up colors.

((rand() % 176) * 80) / 256.0f

I am new to objective c so I can't figure out how this is working. I want to make exact copy of this for Android in Java .

In Java we usually use Random() . How am i suppose to implement this above function using Random r = Random();

Answer 1

In Android, I'd first initialize a variable rand = new Random() . Then I would write your expression as:

rand.nextInt(176) / 3.2f

(Note that 80 / 256.0 == 1 / 3.2 .) I would only assign a value to rand once and reuse the same Random object each time I needed a new color.

After a little back-of-the-envelope work, it seems that your original code is just a fancy way of computing a random float value uniformly distributed between 0 and 55.0f. Thus, a much simpler way of doing the same thing would be:

rand.nextFloat(55)

The only disadvantage of this is that it doesn't resemble the original code very closely (although it will behave the same).

Answer 2

Equivalent one-liner in Java would be:

((new Random().nextInt() % 176) * 80) / 256.0f;

More about the random class on the JavaDoc

Obviously you should not create a new instance of Random each time.

Random r = new Random();
// call r.nextInt() each time you need a new random integer
double color = ((r.nextInt() % 176) * 80) / 256.0f;