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  1. foldl :: (a -> Word8 -> a) -> a -> Bytes -> a

    byteslice Data.Bytes

    Left fold over bytes, non-strict in the accumulator.

  2. foldl :: (a -> b -> a) -> a -> EnumMap k b -> a

    enummapset Data.EnumMap.Lazy

    No documentation available.

  3. foldl :: (a -> b -> a) -> a -> EnumMap k b -> a

    enummapset Data.EnumMap.Strict

    No documentation available.

  4. foldl :: Enum k => (a -> k -> a) -> a -> EnumSet k -> a

    enummapset Data.EnumSet

    No documentation available.

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

    protolude Protolude

    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 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.

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

    protolude Protolude.Partial

    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 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.

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

    ghc-internal GHC.Internal.Data.Foldable

    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 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.

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

    ghc-internal GHC.Internal.Data.List

    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 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.

  9. foldl :: forall a b . (b -> a -> b) -> b -> [a] -> b

    ghc-internal GHC.Internal.Data.OldList

    foldl, 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
    The list must be finite. 
    >>> foldl (+) 0 [1..4]
10

>>> foldl (+) 42 []
42

>>> foldl (-) 100 [1..4]
90

>>> foldl (\reversedString nextChar -> nextChar : reversedString) "foo" ['a', 'b', 'c', 'd']
"dcbafoo"

>>> foldl (+) 0 [1..]
* Hangs forever *

  10. foldl :: forall a b . (b -> a -> b) -> b -> [a] -> b

    ghc-internal GHC.Internal.List

    foldl, 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
    The list must be finite. 
    >>> foldl (+) 0 [1..4]
10

>>> foldl (+) 42 []
42

>>> foldl (-) 100 [1..4]
90

>>> foldl (\reversedString nextChar -> nextChar : reversedString) "foo" ['a', 'b', 'c', 'd']
"dcbafoo"

>>> foldl (+) 0 [1..]
* Hangs forever *

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