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Within LTS Haskell 24.22 (ghc-9.10.3)
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stm-conduit Data.Conduit.Async An operator form of buffer. In general you should be able to replace any use of $$ with $$& and suddenly reap the benefit of concurrency, if your conduits were spending time waiting on each other. The underlying monad must always be an instance of 'MonadBaseControl IO'. If at least one of the two conduits is a CFConduit, it must additionally be a in instance of MonadResource.
>>> CL.sourceList [1,2,3] $$& CL.consume [1,2,3]
It can be combined with $=& and $=. This creates two threads; the first thread produces the list and the second thread does the map and the consume:>>> CL.sourceList [1,2,3] $$& mapC (*2) $= CL.consume [2,4,6]
This creates three threads. The three conduits all run in their own threads:>>> CL.sourceList [1,2,3] $$& mapC (*2) $=& CL.consume [2,4,6]
>>> CL.sourceList [1,2,3] $$& (mapC (*2) $= mapC (+1)) $=& CL.consume [3,5,7]
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stm-conduit Data.Conduit.Async -
stm-conduit Data.Conduit.Async -
stm-conduit Data.Conduit.Async An operator form of buffer'. In general you should be able to replace any use of =$= with =$=& and $$ either with $$& or =$= and runCConduit and suddenly reap the benefit of concurrency, if your conduits were spending time waiting on each other.
>>> runCConduit $ CL.sourceList [1,2,3] =$=& CL.consume [1,2,3]
(
$& ) :: Applicative f => f (a -> b) -> f a -> f bsynthesizer-dimensional Synthesizer.Dimensional.Signal No documentation available.
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bluefin-internal Bluefin.Internal type (:&) :: Effects -> Effects -> Effects
Union of effects -
cabal-install-solver Distribution.Solver.Modular.Version Intersect two version ranges.
(
<&> ) :: Functor f => f a -> (a -> b) -> f bincipit-base Incipit.Base Flipped version of <$>.
(<&>) = flip fmap
Examples
Apply (+1) to a list, a Just and a Right:>>> Just 2 <&> (+1) Just 3
>>> [1,2,3] <&> (+1) [2,3,4]
>>> Right 3 <&> (+1) Right 4
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.&&. ) :: (a -> Bool) -> (a -> Bool) -> a -> Boolliquidhaskell-boot Language.Haskell.Liquid.Misc No documentation available.
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mfsolve Math.MFSolve Make the pairs of expressions on both sides equal. No error is signaled if the equation for one of the sides is Redundant for example in (x, 0) == (y, 0).