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Within LTS Haskell 23.17 (ghc-9.8.4)

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Exact lookup

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

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

   base	Data.List

   (++) 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]
   

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

   base	GHC.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]
   

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

   base	GHC.List

   (++) 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]

   base	GHC.OldList

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

   hedgehog	Hedgehog.Internal.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. type (m :: Symbol) ++ (n :: Symbol) = AppendSymbol m n

   constraints	Data.Constraint.Symbol

   An infix synonym for AppendSymbol.

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

   haskell-gi-base	Data.GI.Base.ShortPrelude

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

   rio	RIO.List

   (++) 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]
   

10. (++) :: [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]
    

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