I've been going through a Haskell tutorial recently and noticed this behaviour when trying some simple Haskell expressions in the interactive ghci
shell:
开发者_运维问答Prelude> 1.1 + 1.1 == 2.2
True
Prelude> 1.1 + 1.1 + 1.1 == 3.3
False
Prelude> 1.1 + 1.1 + 1.1 > 3.3
True
Prelude> 1.1 + 1.1 + 1.1
3.3000000000000003
Does anybody know why that is?
Because 1.1
and 3.3
are floating point numbers. Decimal fractions, such as .1 or .3, are not exactly representable in a binary floating point number. .1 means 1/10. To represent that in binary, where each fractional digit represents 1/2n (1/2, 1/4, 1/8, etc), you would need an infinite number of digits, 0.000110011... repeating infinitely.
This is exactly the same problem as representing, say, 1/3 in base 10. In base 10, you would need an infinite number of digits, .33333... forever, to represent 1/3 exactly. So working in base 10, you usually round, to something like .33. But if you add up three copies of that, you get .99, not 1.
For far more information on the topic, read What Every Computer Scientist Should Know About Floating Point Arithmetic.
For representing rational numbers more precisely in Haskell, you can always use the rational data type, Ratio
; coupled with bignums (arbitrarily large integers, Integer
in Haskell, as opposed to Int
which are fixed size) as the type for numerator and denominator, you can represent arbitrarily precise rational numbers, but at a significantly slower speed than floating point numbers, which are implemented in hardware and optimized for speed.
Floating point numbers are an optimization, for scientific and numerical computation, that trade off precision for high speed, allowing you to perform a very large number of computations in a small time, as long as you are aware of rounding and how it affects your computations.
Because floating-point numbers are not accurate (wikipedia)
You can avoid floating-point errors in Haskell using rational types:
Prelude Data.Ratio> let a = (11 % 10) + (11 % 10) + (11 % 10)
Prelude Data.Ratio> a > (33 % 10)
False
Prelude Data.Ratio> fromRational a
3.3
Of course, you pay a performance penalty for the increased accuracy.
Looks like a typical floating point error issue.
See What is a simple example of floating point/rounding error?
It has to do with the way IEEE floating point numbers work.
1.1 is represented as 1.1000000000000001 in floating point, 3.3 is represented as 3.2999999999999998.
So 1.1 + 1.1 + 1.1 is actually
1.1000000000000001 + 1.1000000000000001 + 1.1000000000000001 = 3.3000000000000003
Which, as you can see is actually larger than 3.2999999999999998.
The usual workaround is to either not evaluate equality, or to check if a number is within the target +/- a small epsilon (which defines the accuracy you need).
Ex: if both are true, then the sum is "equal" to 3.3 (within the allowed error).
1.1 + 1.1 + 1.1 < 3.3 + 1e9
1.1 + 1.1 + 1.1 > 3.3 - 1e9
Few floats can be expressed exactly using IEEE 754 representation, so they will always be a little off.
In general, you shouldn't be comparing floats for equality (for the reasons outlined above). The only reason I can think of is if you want to say "has this value changed?" For example, "if (newscore /= oldscore)" then take some action. That is okay as long as you are not comparing the result of two separate computations to check whether they happen to be equal (because then even mathematically if they are, they might round otherwise).
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