Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
934 views
in Technique[技术] by (71.8m points)

haskell - Why does ghci say that 1.1 + 1.1 + 1.1 > 3.3 is True?

I've been going through a Haskell tutorial recently and noticed this behaviour when trying some simple Haskell expressions in the interactive ghci shell:

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?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...