function-composition

I was reading here , and I noticed that, for example, if I have the following function definitions: a :: Integer->Integer->Integer b :: Integer->Bool The following expression is invalid : (b . a) 2 3 It's quite strange that the functions of the composition must have only one parameter. Is this restriction because some problem in defining the most generic one in Haskell or have some other reason? I'm new to Haskell, so I'm asking maybe useless questions. When you do a 2 3 , you're not applying a to 2 arguments. You're actually applying a to it's only argument, resulting in a function, then take

I'm reading «Scala in Depth» by Joshua Suereth, book that I've bought for the clearly established competency of the author. I'm on page 3 and after a bunch of typos and incoherent formatting (ok, I've become tolerant of these errors) I stumbled upon the following example about a functional approach to solve a very simple scenario. trait Cat trait Bird trait Catch trait FullTummy def catch(hunter: Cat, prey: Bird): Cat with Catch def eat(consumer: Cat with Catch): Cat with FullTummy val story = (catch _) andThen (eat _) story(new Cat, new Bird) I took the example with caution provided it's

Previously asked similar question but somehow I'm not finding my way out, attempting again with another example. The code as a starting point (a bit trimmed) is available at https://ideone.com/zkQcIU . (it has some issue recognizing Microsoft.FSharp.Core.Result type, not sure why) Essentially all operations have to be pipelined with the previous function feeding the result to the next one. The operations have to be async and they should return error to the caller in case an exception occurred. The requirement is to give the caller either result or fault. All functions return a Tuple populated

I was wondering if Haskell keeps track of weather a function is a function composition, i.e would it be possible for me to define a function that does something similar to this?: compositionSplit f.g = (f,g) No, it wouldn't be possible. For example, f1 = (+ 1) . (+ 1) :: Int -> Int is the same function as f2 = subtract 1 . (+ 3) :: Int -> Int and referential transparency demands that equals can be substituted for equals, so if compositionSplit were possible, it would need to produce the same result for f1 and f2 , since that is the same function, yet compositionSplit f1 = ((+ 1), (+1)) and

This question already has an answer here: Feed elements of a tuple to a function as arguments in Haskell? 3 answers Sometimes I have two functions of the form: f :: a -> (b1,b2) h :: b1 -> b2 -> c and I need the composition g. I solve this by changing h to h': h' :: (b1,b2) -> c Can you please show me (if possible) a function m, so that: (h . m . f) == (h' . f) Or another way to deal with such situations. Thanks. What you're looking to do is to take a function that operates on curried arguments, h , and apply it to the result of f , which is a tuple. This process, turning a function of two

This is related to this question: How to do function composition? I noticed that a method reference can be assigned to a variable declared as Function , and so I assume it should have andThen or compose function, and hence I expect that we can compose them directly. But apparently we need to assign them to a variable declared as Function first (or type-cast before invocation) before we can call andThen or compose on them. I suspect I might have some misconception about how this should work. So my questions: Why do we need to type-cast or assign it to a variable first before we can call the