| Re: Strange problem [message #28133 is a reply to message #2626] |
Mon, 26 November 2001 07:00   |
Martin Downing
Messages: 136
Registered: September 1998
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Senior Member |
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Andre,
Joshi is right, this behaviour is due to the lack of precision in
floating point number representation. With your for loop
for i=0., 0.801, 0.1 do print,i
The code execution can more easily be visualised as
i = 0.
while i LE 0.8 do begin
print, i
i = i + 0.1
endwhile
Thus:
IDL> for i=0., 0.8, 0.1 do print,i
0.000000
0.100000
0.200000
0.300000
0.400000
0.500000
0.600000
0.700000
So what was the final value of i?
IDL> print, i
0.800000
Oh, isnt that the value of the upper bound?
IDL> print, i EQ 0.8
0
IDL> print, i - 0.8
5.96046e-008
Clearly not! Slightly more than 0.1 was added each time, so there was a
small excess to i when representing 0.8
So the moral is that you have to be very careful when applying comparison
operators to floating point numbers, one of which is implicitly applied in
the FOR statement. Now you realise the problem, the answer is to be a little
less strict with your comparisons. With FOR loops you can add a small
excess, relative to the increment, to the upper bound:
IDL> for i=0., 0.8001, 0.1 do print,i
0.000000
0.100000
0.200000
0.300000
0.400000
0.500000
0.600000
0.700000
0.800000
Out of interest notice that the final value of "i" is now 0.9:
IDL> print, i
0.900000
> should i be worried?
Well if you write code which depends on floating point numbers having
perfect precision then yes!
If you wanted to compare two floats for equality, you have to rethink what
you mean by "equal", i.e. how exact does this application need the variables
to be?
Relying on doubles is not a robust solution, so instead of writing:
IF a EQ b THEN ...
write
myPrecision = 0.001
IF abs(a-b) LT myPrecision THEN ......
I hope this helps
Martin
"Bhautik Joshi" <nbj@imag.wsahs.nsw.gov.au> wrote in message
news:3C019CFD.B20DB925@imag.wsahs.nsw.gov.au...
>> Anybody know what's going on?
>
> My spin on it:
>
> Well, I think it may be a problem that goes right down to the core, and
> is not just restricted to FOR loops.
>
> Example:
>
> MOO>a=replicate(0.1,100)
> MOO>print, total(a)
> 10.0000
> MOO>print, total(a) - 0.1*100
> 1.90735e-06
>
> argh! where I think the problem may lie is with the binary
> representation of floating point numbers. Now, its been a
> loooooooooooooooooooooooong while since I did computer architecture
> (yes, it is a real subject at uni) but as far as I remember, in floating
> point binary, there is no 'exact' representation for 0.1.
>
> However, lets take 0.5 - which is a quarter of 2; something nice and
> (pardon the expression) base-twoish.
>
> MOO>a=replicate(0.5,100)
> MOO>print, total(a) - 0.5*100
> 0.00000
>
> Lets change the game slightly again:
>
> MOO>a=replicate(0.51,100)
> MOO>print, total(a) - 0.51*100
> -4.95911e-05
>
> should i be worried?
>
>
> --
> /--------------------------------------------------(__)----- ----\
> | nbj@imag.wsahs.nsw.gov.au | phone: 0404032617 |..|--\ -moo |
> | ICQ #: 2464537 | http://cow.mooh.org | |--| |
> \--------------------------------------------------\OO/|| ------/
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