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Within LTS Haskell 24.33 (ghc-9.10.3)

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  1. (+=) :: Annotate ann -> ann -> Annotate ann

    cmdargs System.Console.CmdArgs.Annotate

    Add an annotation to a value.

  2. (+=) :: Annotate ann -> ann -> Annotate ann

    cmdargs System.Console.CmdArgs.Implicit

    Add an annotation to a value.

  3. (++) :: [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]

  4. (++) :: [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]

  5. (++) :: [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]

  6. (++) :: Vector v a => v a -> v a -> v a

    rio RIO.Vector

    No documentation available.

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

    rio RIO.Vector.Boxed

    No documentation available.

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

    rio RIO.Vector.Storable

    No documentation available.

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

    rio RIO.Vector.Unboxed

    No documentation available.

  10. (+=) :: (MonadState s m, Num a) => ASetter' s a -> a -> m ()

    diagrams-lib Diagrams.Prelude

    Modify the target(s) of a Lens', Iso, Setter or Traversal by adding a value. Example: 

    fresh :: MonadState Int m => m Int
fresh = do
id += 1
use id

    >>> execState (do _1 += c; _2 += d) (a,b)
(a + c,b + d)

    >>> execState (do _1.at 1.non 0 += 10) (Map.fromList [(2,100)],"hello")
(fromList [(1,10),(2,100)],"hello")

    (+=) :: (MonadState s m, Num a) => Setter' s a    -> a -> m ()
(+=) :: (MonadState s m, Num a) => Iso' s a       -> a -> m ()
(+=) :: (MonadState s m, Num a) => Lens' s a      -> a -> m ()
(+=) :: (MonadState s m, Num a) => Traversal' s a -> a -> m ()

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