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  1. (++) :: forall (n :: Nat) a (m :: Nat) . Vec n a -> Vec m a -> Vec (n + m) a

    clash-prelude Clash.Explicit.Prelude

    Append two vectors. 

    >>> (1:>2:>3:>Nil) ++ (7:>8:>Nil)
1 :> 2 :> 3 :> 7 :> 8 :> Nil

  2. (++) :: forall (n :: Nat) a (m :: Nat) . Vec n a -> Vec m a -> Vec (n + m) a

    clash-prelude Clash.Explicit.Prelude.Safe

    Append two vectors. 

    >>> (1:>2:>3:>Nil) ++ (7:>8:>Nil)
1 :> 2 :> 3 :> 7 :> 8 :> Nil

  3. (++) :: forall (n :: Nat) a (m :: Nat) . Vec n a -> Vec m a -> Vec (n + m) a

    clash-prelude Clash.Prelude

    Append two vectors. 

    >>> (1:>2:>3:>Nil) ++ (7:>8:>Nil)
1 :> 2 :> 3 :> 7 :> 8 :> Nil

  4. (++) :: forall (n :: Nat) a (m :: Nat) . Vec n a -> Vec m a -> Vec (n + m) a

    clash-prelude Clash.Prelude.Safe

    Append two vectors. 

    >>> (1:>2:>3:>Nil) ++ (7:>8:>Nil)
1 :> 2 :> 3 :> 7 :> 8 :> Nil

  5. (++) :: forall (n :: Nat) a (m :: Nat) . Vec n a -> Vec m a -> Vec (n + m) a

    clash-prelude Clash.Sized.Vector

    Append two vectors. 

    >>> (1:>2:>3:>Nil) ++ (7:>8:>Nil)
1 :> 2 :> 3 :> 7 :> 8 :> Nil

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

    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 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]

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

    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 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]

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

    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 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. type family (xs :: [α]) ++ (ys :: [α]) :: [α]

    fixed-vector-hetero Data.Vector.HFixed.Class

    Concaternation of type level lists.

  10. type family (xs :: [α]) ++ (ys :: [α]) :: [α]

    fixed-vector-hetero Data.Vector.HFixed.TypeFuns

    Concaternation of type level lists.

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