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Programming >> Liberty BASIC language >> maths problem
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maths problem
Post by hammerjit on Apr 21st, 2015, 07:03am

Hi, I am using LBB and when run this code I am not getting "same or more" as a result.

Can someone please guide me.

Code:
hkval = 50000*.60

valueD = hkval/1000

valueC = int(valueD)


if valueC < valueD then
   print "less"
else
    print "same or more"
end if
Re: maths problem
Post by tsh73 on Apr 21st, 2015, 07:23am

You are expecting exact equal of two numbers
Problem is that valueD is floating-point number
(example: you could not exactly represent 1/3 in decimals, you'll get 0.3333333333333. About same way, computer cannot present 1/10 exactly in binary.)
They are stored in computer memory approximately.
Even if you expect it to be 30 it might be something like 30.000000000000001 or 29.9999999999999

The only things that could be guaranteed be exactly equal is integers.
Floating point numbers might be meaningfully compared with given precision.
Code:
precision = 1e-5

a = 1
b = 1.0001

if abs(a-b)<precision then
    print "a and b equal up to ";precision
else
    print "a and b are different"
end if

b = 1.000001
if abs(a-b)<precision then
    print "a and b equal up to ";precision
else
    print "a and b are different"
end if
Re: maths problem
Post by hammerjit on Apr 21st, 2015, 07:34am

so how do I apply it in my example code? is there a possible way?
Re: maths problem
Post by tsh73 on Apr 21st, 2015, 08:46am

Code:
precision = 1e-5

hkval = 50000*.60

valueD = hkval/1000

valueC = int(valueD)


if valueC < valueD-precision then
   print "definitely less"
else
    print "approximately same or more"
end if
Re: maths problem
Post by hammerjit on Apr 21st, 2015, 08:53am

thank you so much....that really helped!!!
grin grin grin
Re: maths problem
Post by Richard Russell on Apr 21st, 2015, 09:09am

on Apr 21st, 2015, 07:34am, hammerjit wrote:
so how do I apply it in my example code? is there a possible way?

What is the problem you are trying to solve? Your code, as it stands, appears to be trying to check whether or not valueD is an integer. It is giving the correct result, because in this case valueD is actually:

Code:
30.00000000000000000173472347597680709441192448139190673828125

(for the reason Anatoly explained).

LBB performs its floating-point calculations to considerably greater accuracy than JB or LB (about 19 significant figures compared with about 16) but small differences of this kind are unavoidable unless you use only integer arithmetic.

Richard.
Re: maths problem
Post by Richard Russell on Apr 21st, 2015, 2:52pm

Here's a good illustration of why you have to be careful. Make a small change to the program: change 0.60 to 0.56 as follows:

Code:
hkval = 50000*.56

valueD = hkval/1000

valueC = int(valueD)


if valueC < valueD then
   print "less"
else
    print "same or more"
end if

Now the program reports "less" when run in JB and "same or more" when run in LBB - in other words the opposite of the earlier results!

Richard.
Re: maths problem
Post by hammerjit on Apr 22nd, 2015, 06:23am

yes Richard, I understand it now. Thanks again