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Within LTS Haskell 24.31 (ghc-9.10.3)
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(
++ ) :: SymVal a => SList a -> SList a -> SList asbv 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]
(
++ ) :: SString -> SString -> SStringsbv 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
type (l :: Symbol)
++ (r :: Symbol) = AppendSymbol l rtype-level-show TypeLevelShow.Utils No documentation available.
(
++ ) :: forall (m :: Type -> Type) a . Monad m => Stream m a -> Stream m a -> Stream m avector-stream Data.Stream.Monadic Concatenate two Streams
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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 slowdownExamples
>>> [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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dimensional Numeric.Units.Dimensional.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 slowdownExamples
>>> [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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distribution-opensuse OpenSuse.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 slowdownExamples
>>> [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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faktory Faktory.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 slowdownExamples
>>> [1, 2, 3] ++ [4, 5, 6] [1,2,3,4,5,6]
>>> [] ++ [1, 2, 3] [1,2,3]
>>> [3, 2, 1] ++ [] [3,2,1]
type family (xs :: [α])
++ (ys :: [α]) :: [α]fixed-vector-hetero Data.Vector.HFixed.Class Concaternation of type level lists.
type family (xs :: [α])
++ (ys :: [α]) :: [α]fixed-vector-hetero Data.Vector.HFixed.TypeFuns Concaternation of type level lists.