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  2. RPG400-L
  3. April 2007

Re: Floating point conversion

callp ceeran0(seed: float: *omit)
Let x = (%dech(float:63:62) * (max-min + 1))

If max = 15 and min = 5, we have
x = (%dech(float:63:62) * (15-5 + 1))
= (%dech(float:63:62) * (11))
= [0,10] after truncation occurs, where [x,y] is the range of numbers x-y, including x and y.

Now, we have
rnd = [0,10] + min
= [0,10] + 5
= [5,15]

That's how I did it:
callp ceeran0(seed: float: *omit)
rnd = (%dech(float:63:62)
* (max-min+1))
+ min

When I used
rnd = (%dech(float:63:62)
* (max-min+1))
+ (min-1)

I got zero results for [1,10], which makes sense.
(max-min+1) = (10-9+1) = 10
(min-1) = (1-1) = 0

if CEERAN0 returns zero, we get
((0 * 10) + 0)
(0 + 0)
0

Your method looks good.
--buck

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