Protolude.Partial

Documentation

head :: [a] -> a

\(\mathcal{O}(1)\). Extract the first element of a list, which must be non-empty.

>>> head [1, 2, 3]
1
>>> head [1..]
1
>>> head []
*** Exception: Prelude.head: empty list

init :: [a] -> [a]

\(\mathcal{O}(n)\). Return all the elements of a list except the last one. The list must be non-empty.

>>> init [1, 2, 3]
[1,2]
>>> init [1]
[]
>>> init []
*** Exception: Prelude.init: empty list

tail :: [a] -> [a]

\(\mathcal{O}(1)\). Extract the elements after the head of a list, which must be non-empty.

>>> tail [1, 2, 3]
[2,3]
>>> tail [1]
[]
>>> tail []
*** Exception: Prelude.tail: empty list

last :: [a] -> a

\(\mathcal{O}(n)\). Extract the last element of a list, which must be finite and non-empty.

>>> last [1, 2, 3]
3
>>> last [1..]
* Hangs forever *
>>> last []
*** Exception: Prelude.last: empty list

foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b

Left-associative fold of a structure, lazy in the accumulator. This is rarely what you want, but can work well for structures with efficient right-to-left sequencing and an operator that is lazy in its left argument.

In the case of lists, foldl, when applied to a binary operator, a starting value (typically the left-identity of the operator), and a list, reduces the list using the binary operator, from left to right:

foldl f z [x1, x2, ..., xn] == (...((z `f` x1) `f` x2) `f`...) `f` xn

Note that to produce the outermost application of the operator the entire input list must be traversed. Like all left-associative folds, foldl will diverge if given an infinite list.

If you want an efficient strict left-fold, you probably want to use foldl' instead of foldl. The reason for this is that the latter does not force the inner results (e.g. z `f` x1 in the above example) before applying them to the operator (e.g. to (`f` x2)). This results in a thunk chain \(\mathcal{O}(n)\) elements long, which then must be evaluated from the outside-in.

For a general Foldable structure this should be semantically identical to:

foldl f z = foldl f z . toList

Examples

The first example is a strict fold, which in practice is best performed with foldl'.

>>> foldl (+) 42 [1,2,3,4]
52

Though the result below is lazy, the input is reversed before prepending it to the initial accumulator, so corecursion begins only after traversing the entire input string.

>>> foldl (\acc c -> c : acc) "abcd" "efgh"
"hgfeabcd"

A left fold of a structure that is infinite on the right cannot terminate, even when for any finite input the fold just returns the initial accumulator:

>>> foldl (\a _ -> a) 0 $ repeat 1
* Hangs forever *

WARNING: When it comes to lists, you always want to use either foldl' or foldr instead.

foldr :: Foldable t => (a -> b -> b) -> b -> t a -> b

Right-associative fold of a structure, lazy in the accumulator.

In the case of lists, foldr, when applied to a binary operator, a starting value (typically the right-identity of the operator), and a list, reduces the list using the binary operator, from right to left:

foldr f z [x1, x2, ..., xn] == x1 `f` (x2 `f` ... (xn `f` z)...)

Note that since the head of the resulting expression is produced by an application of the operator to the first element of the list, given an operator lazy in its right argument, foldr can produce a terminating expression from an unbounded list.

For a general Foldable structure this should be semantically identical to,

foldr f z = foldr f z . toList

Examples

Basic usage:

>>> foldr (||) False [False, True, False]
True

>>> foldr (||) False []
False

>>> foldr (\c acc -> acc ++ [c]) "foo" ['a', 'b', 'c', 'd']
"foodcba"

Infinite structures

⚠️ Applying foldr to infinite structures usually doesn't terminate.

It may still terminate under one of the following conditions:

• the folding function is short-circuiting
• the folding function is lazy on its second argument

Short-circuiting

(||) short-circuits on True values, so the following terminates because there is a True value finitely far from the left side:

>>> foldr (||) False (True : repeat False)
True

But the following doesn't terminate:

>>> foldr (||) False (repeat False ++ [True])
* Hangs forever *

Laziness in the second argument

Applying foldr to infinite structures terminates when the operator is lazy in its second argument (the initial accumulator is never used in this case, and so could be left undefined, but [] is more clear):

>>> take 5 $ foldr (\i acc -> i : fmap (+3) acc) [] (repeat 1)
[1,4,7,10,13]

foldl' :: Foldable t => (b -> a -> b) -> b -> t a -> b

Left-associative fold of a structure but with strict application of the operator.

This ensures that each step of the fold is forced to Weak Head Normal Form before being applied, avoiding the collection of thunks that would otherwise occur. This is often what you want to strictly reduce a finite structure to a single strict result (e.g. sum).

For a general Foldable structure this should be semantically identical to,

foldl' f z = foldl' f z . toList

Since: base-4.6.0.0

foldr' :: Foldable t => (a -> b -> b) -> b -> t a -> b

Right-associative fold of a structure, strict in the accumulator. This is rarely what you want.

Since: base-4.6.0.0

foldr1 :: Foldable t => (a -> a -> a) -> t a -> a

A variant of foldr that has no base case, and thus may only be applied to non-empty structures.

This function is non-total and will raise a runtime exception if the structure happens to be empty.

Examples

Basic usage:

>>> foldr1 (+) [1..4]
10

>>> foldr1 (+) []
Exception: Prelude.foldr1: empty list

>>> foldr1 (+) Nothing
*** Exception: foldr1: empty structure

>>> foldr1 (-) [1..4]
-2

>>> foldr1 (&&) [True, False, True, True]
False

>>> foldr1 (||) [False, False, True, True]
True

>>> foldr1 (+) [1..]
* Hangs forever *

foldl1 :: Foldable t => (a -> a -> a) -> t a -> a

A variant of foldl that has no base case, and thus may only be applied to non-empty structures.

This function is non-total and will raise a runtime exception if the structure happens to be empty.

foldl1 f = foldl1 f . toList

Examples

Basic usage:

>>> foldl1 (+) [1..4]
10

>>> foldl1 (+) []
*** Exception: Prelude.foldl1: empty list

>>> foldl1 (+) Nothing
*** Exception: foldl1: empty structure

>>> foldl1 (-) [1..4]
-8

>>> foldl1 (&&) [True, False, True, True]
False

>>> foldl1 (||) [False, False, True, True]
True

>>> foldl1 (+) [1..]
* Hangs forever *

cycle :: [a] -> [a]

cycle ties a finite list into a circular one, or equivalently, the infinite repetition of the original list. It is the identity on infinite lists.

>>> cycle []
*** Exception: Prelude.cycle: empty list
>>> take 20 $ cycle [42]
[42,42,42,42,42,42,42,42,42,42...
>>> take 20 $ cycle [2, 5, 7]
[2,5,7,2,5,7,2,5,7,2,5,7...

maximum :: (Foldable t, Ord a) => t a -> a

The largest element of a non-empty structure.

This function is non-total and will raise a runtime exception if the structure happens to be empty. A structure that supports random access and maintains its elements in order should provide a specialised implementation to return the maximum in faster than linear time.

Examples

Basic usage:

>>> maximum [1..10]
10

>>> maximum []
*** Exception: Prelude.maximum: empty list

>>> maximum Nothing
*** Exception: maximum: empty structure

WARNING: This function is partial for possibly-empty structures like lists.

Since: base-4.8.0.0

minimum :: (Foldable t, Ord a) => t a -> a

The least element of a non-empty structure.

This function is non-total and will raise a runtime exception if the structure happens to be empty. A structure that supports random access and maintains its elements in order should provide a specialised implementation to return the minimum in faster than linear time.

Examples

Basic usage:

>>> minimum [1..10]
1

>>> minimum []
*** Exception: Prelude.minimum: empty list

>>> minimum Nothing
*** Exception: minimum: empty structure

WARNING: This function is partial for possibly-empty structures like lists.

Since: base-4.8.0.0

(!!) :: [a] -> Int -> a infixl 9

List index (subscript) operator, starting from 0. It is an instance of the more general genericIndex, which takes an index of any integral type.

>>> ['a', 'b', 'c'] !! 0
'a'
>>> ['a', 'b', 'c'] !! 2
'c'
>>> ['a', 'b', 'c'] !! 3
*** Exception: Prelude.!!: index too large
>>> ['a', 'b', 'c'] !! (-1)
*** Exception: Prelude.!!: negative index

sum :: (Foldable t, Num a) => t a -> a

The sum function computes the sum of the numbers of a structure.

Examples

Basic usage:

>>> sum []
0

>>> sum [42]
42

>>> sum [1..10]
55

>>> sum [4.1, 2.0, 1.7]
7.8

>>> sum [1..]
* Hangs forever *

Since: base-4.8.0.0

product :: (Foldable t, Num a) => t a -> a

The product function computes the product of the numbers of a structure.

Examples

Basic usage:

>>> product []
1

>>> product [42]
42

>>> product [1..10]
3628800

>>> product [4.1, 2.0, 1.7]
13.939999999999998

>>> product [1..]
* Hangs forever *

Since: base-4.8.0.0

fromJust :: HasCallStack => Maybe a -> a

The fromJust function extracts the element out of a Just and throws an error if its argument is Nothing.

Examples

Basic usage:

>>> fromJust (Just 1)
1

>>> 2 * (fromJust (Just 10))
20

>>> 2 * (fromJust Nothing)
*** Exception: Maybe.fromJust: Nothing
...

read :: Read a => String -> a

The read function reads input from a string, which must be completely consumed by the input process. read fails with an error if the parse is unsuccessful, and it is therefore discouraged from being used in real applications. Use readMaybe or readEither for safe alternatives.

>>> read "123" :: Int
123

>>> read "hello" :: Int
*** Exception: Prelude.read: no parse