CODEWITHSUNDEEP
haskell
Srinivas
Srinivas

Reputation: 2098

Haskell function return values

what is the difference between

dotEx1 = map(+3) . filter (>100)

and

dotEx1 xs = map(+3) . filter (>100) xs

since 

myFilter xs = filter (>100) xs

and 

myFilter = filter (>100)

are the same why isn't 

dotEx1 = map(+3) . filter (>100)

and 

dotEx1 xs = map(+3) . filter (>100) xs

the same?

Upvotes: 1

Views: 98

Answers (3)

Taren
Taren

Reputation: 684

The dot operator has low precedence because you want to partially apply functions. That is, 

map (+3) . filter (>100)

is read as

(map (+3)) . (filter (>100))

By extension you get

dotEx1 xs = (map (+3)) . (filter (>100) xs)

instead of 

dotEx1 xs = (map (+3) . filter (>100)) xs

A more readable version that also works:

dotEx1 xs = map (+3) . filter (>100) $ xs

Upvotes: 3

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 477641

The dot operator has signature:

(.) :: (b -> c) -> (a -> b) -> a -> c

here the operator is used infix. So in your first statement you have actually written:

dotEx1 = (.) (map (+3)) (filter (>100))

And that makes sense since filter (>100) has signature: (Num n,Ord n) => [n] -> [n] and map (+3) has signature Num n => [n] -> [n]. If you however write:

dotEx1 xs = (.) (map (+3)) (filter (>100) xs)

then filter (>100) xs has signature (Num n,Ord n) => [n] and so this is a value (not a function, or perhaps a function with no arguments). So the dot operator cannot be used (the types do not match).

Informally the dot operator should be given two functions f and g and it generates a function where f is applied after g is applied on the input. But g thus has to be a function taking one argument.

Upvotes: 4

Chad Gilbert
Chad Gilbert

Reputation: 36385

The . function is defined as this:

(.) :: (b -> c) -> (a -> b) -> a -> c

The function composed by the . operator must accept an argument. Therefore,

dotEx1 = map(+3) . filter (>100)

Is the same as

dotEx1 xs = (map(+3) . filter (>100)) xs

Upvotes: 7

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