Hi all
rects*.graphics.beginFill(Math.floor(Math.random( )*0xFFFFFF), 1);
I can’t find , what floor and Math.random( ) do to this code.I have a reference .
Thanks advance .
Hey there, kv79. Math.floor() and Math.random() are being used here to create a random color to use as the fill color. They’re done in two steps:
Math.random() * 0xFFFFFF
This creates a random color/number. It does this by calling Math.random(), which will return a number value from 0-1. That number is then multiplied by 0xFFFFFF, making the possible value range 0x000000 to 0xFFFFFF.
Because 0xFFFFFF is multiplied by a decimal value, it’s possible the end result isn’t a valid color (it’d be like trying to say 0xFFFFFF.667) Hexidecimal values are integral, so to make sure of this, Math.floor() is used. Math.floor() takes a value (in this case the random color just created) and rounds it down to the nearest integer.
How can Hexidecimal number can be multiplied with number between 0-1 ?
Decimal * Hexidecimal => abstract .
well all numbers are stored binary, so u don’t have to worry
Yeah, as far as Flash is concerned, 0x00001B is the same thing as saying 27.
Instead of using Math.floor(), use the uint(), it’s faster, and Math floor actually produces a Number, not an Integer. Because of accuracy problems with floating values, Math.floor(4.7458) could return 4.000000008758
Therefore use uint(), it just cuts away everything behind the coma.
uint(Math.random()*0xFFFFFF)
But actually I think that it’s unnecessary to round it, flash does that by itself, because beginFill only takes uint’s (from the Flash language reference):
beginFill(color:uint, alpha:Number = 1.0)
So to sum up, your final code could be:
rects*.graphics.beginFill(Math.random()*0xFFFFFF);
But the other suggestions will work just fine, but this is faster and easier(for further coding)
By the way, avoid setting alpha to 1, it’s the default value:thumb2:
It’s a uint typed parameter anyway, the Math.floor() is unnecessary.
Thanks to all .
