why this logical expression is wrong?

Hi,
why this statment is wrong,
>> sind(30)==0.5
ans =
0

4 Commenti

Which release of MATLAB are you using?
Are you using the built-in sind function or one you've written or downloaded? In other words, what does this command return?
which -all sind
I'm using matlab 2016 a and also i'm using built-in function.
built-in (C:\Program Files\MATLAB\R2016a\toolbox\matlab\elfun\@double\sind) % double method
built-in (C:\Program Files\MATLAB\R2016a\toolbox\matlab\elfun\@single\sind) % single method
C:\Program Files\MATLAB\R2016a\toolbox\distcomp\parallel\@codistributed\sind.m % codistributed method
C:\Program Files\MATLAB\R2016a\toolbox\distcomp\gpu\@gpuArray\sind.m % gpuArray method
I seem to remember seeing a sind bug in the bug reports a couple of releases ago, but I cannot find that information now.
I think it's not a bug, because it depends on the method of approximation. That's an interesting question how developers of sind() function fixed sind(30). Whether there is an if statement in the code of sind() in order to set sind(30) exactly 1/2. :-)

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 Risposta accettata

Peter Jarosi
Peter Jarosi il 18 Lug 2019
Modificato: Peter Jarosi il 18 Lug 2019

1 voto

because of the accuracy problem of a floating-point system
>> format longE
>> sind(30)
ans =
4.999999999999999e-01
Never, ever compare two floating-point numbers with == operand!
See also:

11 Commenti

In current releases,
>> sind(30) - 0.5
ans =
0
Peter Jarosi
Peter Jarosi il 18 Lug 2019
Modificato: Peter Jarosi il 18 Lug 2019
I use a little bit older version:
>> format longE
>> sind(30) - 0.5
ans =
-5.551115123125783e-17
It's my fault. Anyway, it's a programming principle to avoid floating-point numbers comparison.
>> help sind
sind Sine of argument in degrees.
sind(X) is the sine of the elements of X, expressed in degrees.
For integers n, sind(n*180) is exactly zero, whereas sin(n*pi)
reflects the accuracy of the floating point value of pi.
tanks
but I try this before in this way:
format long
>> sind(30)
ans =
0.500000000000000
as you can see I get same result!
format long rounds off to 15 decimal places instead of all 16. You need to subtract 0.5 to see the difference between the sind(30) result and 0.5
Sometimes we can be lucky with floating-point numbers (depending on the function or the version of our system) but never sure! That's why it's better to compare two floating-point numbers within a tolerance:
tolerance = 1e-10;
abs(a - b) < tolerance;
you're right.
thank you
You're welcome! Thank you for accepting my answer.
I voted your question. It's a million dollar question, and the problem is general, not Matlab's fault.
The result of sind(30) should be exactly 1/2, but it appears that in some older releases it was not exactly that.
It is generally better to not compare for floating point equality except when you know that the exact bit-pattern occurs somewhere. For example, it is fair to test
A = min(B);
B == A
because you know that A will be a bit-for-bit copy of one of the values that is in B. (Well, except for some obscure cases involving non-default nan values... and in those cases, nan == nan is always false anyhow.)

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Più risposte (1)

Steven Lord
Steven Lord il 19 Lug 2019

2 voti

This is a bug. See Bug Report 1839169.

1 Commento

How did you fix it? Did you use a better approximation method (for instance longer Taylor series) or just put an if statement in the code of function sind()? :-)
(I'm joking)

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