# Hoogle Search

Within LTS Haskell 22.29 (ghc-9.6.6)

*Note that Stackage only displays results for the latest LTS and Nightly snapshot.*
Learn more.

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

The first example is a strict fold, which in practice is best performed with foldl'.**Examples**>>> 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.foldl :: Foldable t => (b -> a -> b) -> b -> t a -> bamazonka-core Amazonka.Prelude 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

The first example is a strict fold, which in practice is best performed with foldl'.**Examples**>>> 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.foldl :: Foldable t => (b -> a -> b) -> b -> t a -> bhedgehog Hedgehog.Internal.Prelude 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

The first example is a strict fold, which in practice is best performed with foldl'.**Examples**>>> 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.foldl :: Foldable t => (b -> a -> b) -> b -> t a -> bghc GHC.Prelude.Basic 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

The first example is a strict fold, which in practice is best performed with foldl'.**Examples**>>> 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.foldl :: Foldable t => (b -> a -> b) -> b -> t a -> bCabal-syntax Distribution.Compat.Prelude 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

The first example is a strict fold, which in practice is best performed with foldl'.**Examples**>>> 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.foldl :: Foldable t => (b -> a -> b) -> b -> t a -> bghc-lib-parser GHC.Prelude.Basic 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

The first example is a strict fold, which in practice is best performed with foldl'.**Examples**>>> 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.foldl :: Foldable t => (b -> a -> b) -> b -> t a -> bnumhask NumHask.Prelude 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

The first example is a strict fold, which in practice is best performed with foldl'.**Examples**>>> 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.foldl :: Foldable t => (b -> a -> b) -> b -> t a -> bnumhask NumHask.Prelude 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

The first example is a strict fold, which in practice is best performed with foldl'.**Examples**>>> 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.foldl :: (a -> b -> a) -> a -> [b] -> aprelude-compat Prelude2010 No documentation available.

foldl :: Foldable t => (b -> a -> b) -> b -> t a -> bmixed-types-num Numeric.MixedTypes.PreludeHiding 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

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

Page 1 of many | Next