Fully expanding your program might help. One of the great things about Haskell is equational reasoning: if two things have been declared equal, you can substitute one for the other. First, let's desugar that do notation to the equivalent bind chain: multWithLog = logNumber 3 >>= \a -> logNumber 5 >>= \b -> return (a*b) Evaluate the logNumber and return calls to normal form from their definitions, also considering the monoid definitions of (++) and mempty for lists: multWithLog = Writer (3, ["Got number: 3"]) >>= \a -> Writer (5, ["Got number: 5"]) >>= \b -> Writer (a*b, []) Now, refer to the definition of (>>=) for Writer (as shown in LYAH): (Writer (x, v)) >>= f = let (Writer (y, v')) = f x in Writer (y, v `mappend` v') Rewritten as a lamba (and replacing `mappend` with (++) for brevity), this becomes: \(Writer (x, v)) f -> let (Writer (y, v')) = f x in Writer (y, v++v') It's no longer an infix function, so we'll have to shift things around a little. Here's what it expands to: multWithLog = (\(Writer (x, v)) f -> -- \ let (Writer (y, v')) = f x -- | bind in Writer (y, v++v')) -- / (Writer (3, ["Got number: 3"])) -- Writer (x, v) (\a -> -- f (\(Writer (x2, v2)) f2 -> -- \ let (Writer (y2, v2')) = f2 x2 -- | bind in Writer (y2, v2++v2')) -- / (Writer (5, ["Got number: 5"])) -- Writer (x2, v2) (\b -> Writer (a*b, [])) -- f2 ) -- (end f) Now it's just a matter of simplification. Let's start by eliminating the first argument of each bind, i.e. (Writer (x,v)), by substituting with concrete values. multWithLog = (\f -> -- \ partially let (Writer (y, v')) = f 3 -- | applied in Writer (y, ["Got number: 3"]++v')) -- / bind (\a -> -- f (\f2 -> -- \ partially let (Writer (y2, v2')) = f2 5 -- | applied in Writer (y2, ["Got number: 5"]++v2')) -- / bind (\b -> Writer (a*b, [])) -- f2 ) -- (end f) Substitute f2, and eliminate both \f2 and \b in the same way: multWithLog = (\f -> let (Writer (y, v')) = f 3 in Writer (y, ["Got number: 3"]++v')) (\a -> let (Writer (y2, v2')) = Writer (a*5, []) -- applied f2 in Writer (y2, ["Got number: 5"]++v2') ) With a full match on the inner let block, that can also be eliminated: multWithLog = (\f -> let (Writer (y, v')) = f 3 in Writer (y, ["Got number: 3"]++v')) (\a -> Writer (a*5, ["Got number: 5"]++[])) I'll forego the last few substitutions. If it's still not clear how you get to the final output, you should run through them yourself. You'll eventually reach a static definition of the result -- which shouldn't really be surprising, since there were no arguments to this "function" and its type doesn't contain -> anywhere. :) On Thu, Jan 26, 2017 at 1:34 PM, Olumide <50295@web.de> wrote:
On 26/01/17 16:02, David McBride wrote:
runWriter multWithLogTuple
((3,5,10),["Got number: 3","Got number: 5"])
On second thoughts I don't think I understand how the logs are concatenated. I was expecting (15,["Got number: 15") in the original example.
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