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  1. (++) :: [a] -> [a] -> [a]

    cabal-install-solver Distribution.Solver.Compat.Prelude

    (++) appends two lists, i.e., 

    [x1, ..., xm] ++ [y1, ..., yn] == [x1, ..., xm, y1, ..., yn]
[x1, ..., xm] ++ [y1, ...] == [x1, ..., xm, y1, ...]
    If the first list is not finite, the result is the first list.

    Performance considerations

    This function takes linear time in the number of elements of the first list. Thus it is better to associate repeated applications of (++) to the right (which is the default behaviour): xs ++ (ys ++ zs) or simply xs ++ ys ++ zs, but not (xs ++ ys) ++ zs. For the same reason concat = foldr (++) [] has linear performance, while foldl (++) [] is prone to quadratic slowdown

    Examples

    >>> [1, 2, 3] ++ [4, 5, 6]
[1,2,3,4,5,6]

    >>> [] ++ [1, 2, 3]
[1,2,3]

    >>> [3, 2, 1] ++ []
[3,2,1]

  2. type family (as :: [k]) ++ (bs :: [k]) :: [k]

    generic-lens-core Data.Generics.Product.Internal.HList

    No documentation available.

  3. type family (es :: [Effect]) ++ (es' :: [Effect]) :: [Effect]

    heftia Control.Monad.Hefty

    No documentation available.

  4. (++) :: [a] -> [a] -> [a]

    ihaskell IHaskellPrelude

    (++) appends two lists, i.e., 

    [x1, ..., xm] ++ [y1, ..., yn] == [x1, ..., xm, y1, ..., yn]
[x1, ..., xm] ++ [y1, ...] == [x1, ..., xm, y1, ...]
    If the first list is not finite, the result is the first list.

    Performance considerations

    This function takes linear time in the number of elements of the first list. Thus it is better to associate repeated applications of (++) to the right (which is the default behaviour): xs ++ (ys ++ zs) or simply xs ++ ys ++ zs, but not (xs ++ ys) ++ zs. For the same reason concat = foldr (++) [] has linear performance, while foldl (++) [] is prone to quadratic slowdown

    Examples

    >>> [1, 2, 3] ++ [4, 5, 6]
[1,2,3,4,5,6]

    >>> [] ++ [1, 2, 3]
[1,2,3]

    >>> [3, 2, 1] ++ []
[3,2,1]

  5. (++) :: [a] -> [a] -> [a]

    incipit-base Incipit.Base

    (++) appends two lists, i.e., 

    [x1, ..., xm] ++ [y1, ..., yn] == [x1, ..., xm, y1, ..., yn]
[x1, ..., xm] ++ [y1, ...] == [x1, ..., xm, y1, ...]
    If the first list is not finite, the result is the first list.

    Performance considerations

    This function takes linear time in the number of elements of the first list. Thus it is better to associate repeated applications of (++) to the right (which is the default behaviour): xs ++ (ys ++ zs) or simply xs ++ ys ++ zs, but not (xs ++ ys) ++ zs. For the same reason concat = foldr (++) [] has linear performance, while foldl (++) [] is prone to quadratic slowdown

    Examples

    >>> [1, 2, 3] ++ [4, 5, 6]
[1,2,3,4,5,6]

    >>> [] ++ [1, 2, 3]
[1,2,3]

    >>> [3, 2, 1] ++ []
[3,2,1]

  6. (++) :: SymVal a => SList a -> SList a -> SList a

    sbv Data.SBV.List

    Append two lists. 

    >>> sat $ \x y z -> length x .== 5 .&& length y .== 1 .&& x ++ y ++ z .== [1 .. 12]
Satisfiable. Model:
s0 =      [1,2,3,4,5] :: [Integer]
s1 =              [6] :: [Integer]
s2 = [7,8,9,10,11,12] :: [Integer]

  7. (++) :: SString -> SString -> SString

    sbv Data.SBV.String

    Short cut for concat. 

    >>> sat $ \x y z -> length x .== 5 .&& length y .== 1 .&& x ++ y ++ z .== "Hello world!"
Satisfiable. Model:
s0 =  "Hello" :: String
s1 =      " " :: String
s2 = "world!" :: String

  8. type (l :: Symbol) ++ (r :: Symbol) = AppendSymbol l r

    type-level-show TypeLevelShow.Utils

    No documentation available.

  9. (++) :: forall (m :: Type -> Type) a . Monad m => Stream m a -> Stream m a -> Stream m a

    vector-stream Data.Stream.Monadic

    Concatenate two Streams

  10. (++) :: [a] -> [a] -> [a]

    calligraphy Calligraphy.Prelude

    (++) appends two lists, i.e., 

    [x1, ..., xm] ++ [y1, ..., yn] == [x1, ..., xm, y1, ..., yn]
[x1, ..., xm] ++ [y1, ...] == [x1, ..., xm, y1, ...]
    If the first list is not finite, the result is the first list.

    Performance considerations

    This function takes linear time in the number of elements of the first list. Thus it is better to associate repeated applications of (++) to the right (which is the default behaviour): xs ++ (ys ++ zs) or simply xs ++ ys ++ zs, but not (xs ++ ys) ++ zs. For the same reason concat = foldr (++) [] has linear performance, while foldl (++) [] is prone to quadratic slowdown

    Examples

    >>> [1, 2, 3] ++ [4, 5, 6]
[1,2,3,4,5,6]

    >>> [] ++ [1, 2, 3]
[1,2,3]

    >>> [3, 2, 1] ++ []
[3,2,1]

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