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

   rio	RIO.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. (++) :: Vector v a => v a -> v a -> v a

   rio	RIO.Vector

   No documentation available.

3. (++) :: Vector a -> Vector a -> Vector a

   rio	RIO.Vector.Boxed

   No documentation available.

4. (++) :: Storable a => Vector a -> Vector a -> Vector a

   rio	RIO.Vector.Storable

   No documentation available.

5. (++) :: Unbox a => Vector a -> Vector a -> Vector a

   rio	RIO.Vector.Unboxed

   No documentation available.

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

   Cabal-syntax	Distribution.Compat.Prelude

   No documentation available.

7. type family (xs :: [k]) ++ (ys :: [k]) :: [k]

   relude	Relude.Extra.Type

   Concatenates type-level lists.

   >>> :kind! '[ 'Just 5, 'Nothing] ++ '[ 'Just 3, 'Nothing, 'Just 1]
   '[ 'Just 5, 'Nothing] ++ '[ 'Just 3, 'Nothing, 'Just 1] :: [Maybe
   Natural]
   = '[ 'Just 5, 'Nothing, 'Just 3, 'Nothing, 'Just 1]
   

   >>> :kind! '[] ++ '[ 'Just 3, 'Nothing, 'Just 1]
   '[] ++ '[ 'Just 3, 'Nothing, 'Just 1] :: [Maybe Natural]
   = '[ 'Just 3, 'Nothing, 'Just 1]
   

   # 91 "srcReludeExtra/Type.hs"

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

   relude	Relude.List.Reexport

   (++) 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]
   

9. data ((b :: [a]) ++ (c :: [a])) (d :: [a])

   first-class-families	Fcf

   List catenation.

   Example

   >>> data Example where Ex :: a -> Example  -- Hide the type of examples to avoid brittleness in different GHC versions
   
   >>> :kind! Ex (Eval ([1, 2] ++ [3, 4]) :: [Natural])
   Ex (Eval ([1, 2] ++ [3, 4]) :: [Natural]) :: Example
   = Ex [1, 2, 3, 4]
   

10. data ((b :: [a]) ++ (c :: [a])) (d :: [a])

    first-class-families	Fcf.Data.List

    List catenation.

    Example

    >>> data Example where Ex :: a -> Example  -- Hide the type of examples to avoid brittleness in different GHC versions
    
    >>> :kind! Ex (Eval ([1, 2] ++ [3, 4]) :: [Natural])
    Ex (Eval ([1, 2] ++ [3, 4]) :: [Natural]) :: Example
    = Ex [1, 2, 3, 4]
    

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