CODEWITHSUNDEEP
haskell
Nick
Nick

Reputation: 9061

How to compose functions without parenthesis in Haskell?

I am trying to figure out how many digits the 500th Fibonacci number is:

fib n = fibs (0,1) !! n where 
    fibs (a,b) = a:fibs(b,(a+b))
length (show (fib 500))

This does work. Is there a way to get rid of the parenthesis in the third line? Say length . show . fib 500 or something alike?

This dollar-form works too:

length $ show $ fib 500

Generally, what is the recommended way in this situation?

Upvotes: 2

Views: 302

Answers (3)

Lee Duhem
Lee Duhem

Reputation: 15121

You almost got it:

length . show . fib $ 500

You can define a function in pointless style for this

lengthOfFib = length . show . fib

then use it like this

lengthOfFib 500

Upvotes: 5

mhwombat
mhwombat

Reputation: 8136

I would normally use dots to compose all of the functions except the last one. For example:

length . show $ fib 500
this . that . theOther . length . show $ fib 500

You could also write it as:

(length . show . fib) 500
(this . that . theOther . length . show . fib) 500

Upvotes: 3

Jon Purdy
Jon Purdy

Reputation: 55069

Typical style:

length . show $ fib 500

Only compositions:

length . show . fib $ 500

One advantage to preferring composition is that you can easily pull out common subfunctions. Maybe you need length . show somewhere else, so you can trivially move it elsewhere and give it a name; show . fib is probably less useful, though. :)

What you said with:

length . show . fib 500

Is the same as:

\x -> length (show (fib 500)) x

Which the compiler sees as nonsensical because you’re using Int like a function.

Upvotes: 11

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