Relative to the fish operator, Monads satisfy associativity.
(h >=> g) >=> f = h >=> ( g >=> f)
This translated to bind (with lambda expressions) looks like,
\a -> h a >>=(\b -> g b >>= \c -> f c) =
\a ->(h a >>= \b -> g b)>>= \c -> f c
which means the following equation is unambiguous
( a -> h a >>= \b -> g b >>= \c -> f c ) = h >=> g >=> f
This is a nice way to understand Monadic composition.
However not all Monadic code keeps the bound variables to the lambdas separated. For example,
[1,2] >>= \n -> ['a','b'] >>= \ch -> return (n,ch) =
[(1,'a'),(1,'b'),(2,'a'),(2,'b')]
The "n" in the return was obtained from the top lambda.
More generally,
a -> g a >>= \b -> f a b
f depends on both a and b in the above. Defining the above in terms of f, g, and (>=>) gives
\a -> (\x -> g a) >=> f a
Which I don't understand very well. It doesn't match the above equation I showed very well. I see fish as the fundamental notion here, and I'm trying to understand how it interacts with typical Monads I write. I would like to understand the above better.
One way of approaching this is by trying to obtain meaning from List expression syntax
[ (n,ch) | n <- [1, 2], ch <- ['a', 'b'] ]
I think this implies a kind of symmetry.
Are there any nice symmetries connecting lambda expressions and Monads? Or am I reading too much into this? Is my fear of highly nested lambda expressions wrong or reasonable?
No, there aren't any restrictions. Once you've bound a lambda you can do anything . This is one of the reasons of Applicative
being preferrable to Monad
because it's weaker (and hence gives you stronger free-theorem restrictions).
( [1,2] >>= \n -> "ab" >>= \ch -> return (n,ch) )
≡ (,) <$> [1,2] <*> "ab"
≡ liftA2 (,) [1,2] "ab"
≈ liftA2 (flip (,)) "ab" [1,2]
The last is actually not a proper equation; the applicative laws only guarantee that the values will be the same for these expressions See comments , but the structure can and will be different.
Prelude Control.Applicative> liftA2 (,) [1,2] "ab"
[(1,'a'),(1,'b'),(2,'a'),(2,'b')]
Prelude Control.Applicative> liftA2 (flip (,)) "ab" [1,2]
[(1,'a'),(2,'a'),(1,'b'),(2,'b')]
Addressing your edit, in which you consider how to write...
\\a -> ga >>= \\b -> fab
... using (>=>)
, nothing is actually lost in that case. It is helpful to take a step back and consider exactly how (>=>)
can be converted into (>>=)
and vice-versa:
f >=> g = \x -> f x >>= g
m >>= f = (const m >=> f) () -- const x = \_ -> x
In the second equation, which is the one related to your concerns, we turn the first argument to (>>=)
into a function which can be passed to (>=>)
by using const
. As const m >=> f
is a functions which ignores its argument, we just pass ()
as a dummy argument in order to recover (>>=)
.
That being so, your example can be rewritten by using the second equation:
\a -> g a >>= \b -> f a b
\a -> (const (g a) >=> \b -> f a b) ()
\a -> (const (g a) >=> f a) ()
Which, except for the added trick of supplying a dummy ()
, is what you had obtained in your question.
An additional idea to your question: Monads are most general in the sense that effects can depend on inputs. A monadic computation m
that takes input a
and produces output b
can be written as a -> mb
. As this is a function, we (can) define such computations using lambdas, which can naturally span fo the right. But this generality complicates analyzing computations as your \\a -> ga >>= \\b -> fab
.
For arrows (which occupy the space between applicative functors and monads) the situation is somewhat different. For a general arrow, the input must be explicit - an arrow computation arr
has the general type of arr ab
. And so an input that spans "forward" in an arrow computation must be explicitly threaded there using Arrow primitives.
To expand your example
{-# LANGUAGE Arrows #-}
import Control.Arrow
bind2 :: (Monad m) => (a -> m b) -> (a -> b -> m c) -> a -> m c
bind2 g f = \a -> g a >>= \b -> f a b
to arrows: Function f
must now take a pair as its input (because arrows are defined as accepting one input value). Using the arrow do
notation we can express it as
bind2A :: (Arrow arr) => arr a b -> arr (a, b) c -> arr a c
bind2A g f = proc a -> do
b <- g -< a
c <- f -< (a, b)
returnA -< c
Or even simpler using the Arrow
primitives:
bind2A' :: (Arrow arr) => arr a b -> arr (a, b) c -> arr a c
bind2A' g f = (returnA &&& g) >>> f
Graphically:
--------------->[ ]
\ [ f ]---->
\-->[ g ]-->[ ]
Being less general, arrows allow to infer more information about a circuit before it's actually executed. A nice reading is Understanding arrows , which describes the original motivation behind them - to construct parsers that can can avoid space leaks by having static and dynamic parts.
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