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Within LTS Haskell 24.52 (ghc-9.10.3)

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  1. foldlDeque :: forall (v :: Type -> Type -> Type) a m acc . (MVector v a, PrimMonad m) => (acc -> a -> m acc) -> acc -> Deque v (PrimState m) a -> m acc

    rio RIO.Deque

    Fold over a Deque, starting at the beginning. Does not modify the Deque.

  2. foldrDeque :: forall (v :: Type -> Type -> Type) a m acc . (MVector v a, PrimMonad m) => (a -> acc -> m acc) -> acc -> Deque v (PrimState m) a -> m acc

    rio RIO.Deque

    Fold over a Deque, starting at the end. Does not modify the Deque.

  3. freezeDeque :: (Vector v a, PrimMonad m) => Deque (Mutable v) (PrimState m) a -> m (v a)

    rio RIO.Deque

    Yield an immutable copy of the underlying mutable vector. The difference from dequeToVector is that the the copy will be performed with a more efficient memcpy, rather than element by element. The downside is that the resulting vector type must be the one that corresponds to the mutable one that is used in the Deque.

    Example

    >>> :set -XTypeApplications
    
    >>> import qualified RIO.Vector.Unboxed as U
    
    >>> d <- newDeque @U.MVector @Int
    
    >>> mapM_ (pushFrontDeque d) [0..10]
    
    >>> freezeDeque @U.Vector d
    [10,9,8,7,6,5,4,3,2,1,0]
    

  4. getDequeSize :: forall m (v :: Type -> Type -> Type) a . PrimMonad m => Deque v (PrimState m) a -> m Int

    rio RIO.Deque

    O(1) - Get the number of elements that is currently in the Deque

  5. newDeque :: forall (v :: Type -> Type -> Type) a m . (MVector v a, PrimMonad m) => m (Deque v (PrimState m) a)

    rio RIO.Deque

    Create a new, empty Deque

  6. popBackDeque :: forall (v :: Type -> Type -> Type) a m . (MVector v a, PrimMonad m) => Deque v (PrimState m) a -> m (Maybe a)

    rio RIO.Deque

    Pop the first value from the end of the Deque

  7. popFrontDeque :: forall (v :: Type -> Type -> Type) a m . (MVector v a, PrimMonad m) => Deque v (PrimState m) a -> m (Maybe a)

    rio RIO.Deque

    Pop the first value from the beginning of the Deque

  8. pushBackDeque :: forall (v :: Type -> Type -> Type) a m . (MVector v a, PrimMonad m) => Deque v (PrimState m) a -> a -> m ()

    rio RIO.Deque

    Push a new value to the end of the Deque

  9. pushFrontDeque :: forall (v :: Type -> Type -> Type) a m . (MVector v a, PrimMonad m) => Deque v (PrimState m) a -> a -> m ()

    rio RIO.Deque

    Push a new value to the beginning of the Deque

  10. isSubsequenceOf :: Eq a => [a] -> [a] -> Bool

    rio RIO.List

    The isSubsequenceOf function takes two lists and returns True if all the elements of the first list occur, in order, in the second. The elements do not have to occur consecutively. isSubsequenceOf x y is equivalent to x `elem` (subsequences y). Note: isSubsequenceOf is often used in infix form.

    Examples

    >>> "GHC" `isSubsequenceOf` "The Glorious Haskell Compiler"
    True
    
    >>> ['a','d'..'z'] `isSubsequenceOf` ['a'..'z']
    True
    
    >>> [1..10] `isSubsequenceOf` [10,9..0]
    False
    
    For the result to be True, the first list must be finite; for the result to be False, the second list must be finite:
    >>> [0,2..10] `isSubsequenceOf` [0..]
    True
    
    >>> [0..] `isSubsequenceOf` [0,2..10]
    False
    
    >>> [0,2..] `isSubsequenceOf` [0..]
    * Hangs forever*
    

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