Hi, Thanks all for your good help. I was caught up in sequential thinking about monads so much so that I treated the lambda expressions as separate functions rather than a nested big one.
That clears up a lot of nagging doubts. Cheers, Matt. On 20 Jul 2013, at 00:18, Rogan Creswick <cresw...@gmail.com> wrote: > On Fri, Jul 19, 2013 at 3:58 PM, Matt Ford <m...@dancingfrog.co.uk> wrote: >> Hi, >> >> Thanks for the help. >> >> I thought >>= was left associative? It seems to be in the examples from >> Learn You A Haskell. >> >> I tried to use the associative law to bracket from the right but it didn't >> like that either... >> >> [1,2] >>= (\x -> (\n -> [3,4])) x >>= \m -> return (n,m)) > > I think the issue is that you need to first take into account the lambdas > *then* use what you know about the properties of (>>=). > > I found this stackoverflow answer helpful > (http://stackoverflow.com/a/11237469) > > "The rule for lambdas is pretty simple: the body of the lambda extends as far > to the right as possible without hitting an unbalanced parenthesis." > > So, the first lambda runs to the end of the expression: > > [1,2] >>= (\n -> [3,4] >>= \m -> return (n,m)) > > Now, there is still a lambda nested inside the first lambda: \m -> return > (n,m) > > [1,2] >>= (\n -> [3,4] >>= (\m -> return (n,m))) > > You violated the implied grouping that these new parentheses make explicit > when you tried to apply the associative law above. > > Timon's post continues from this point to show the full deconstruction. > > --Rogan > >> >> Any thoughts? >> >> Matt >> >> On 19 Jul 2013, at 23:35, Rogan Creswick <cresw...@gmail.com> wrote: >> >>> On Fri, Jul 19, 2013 at 3:23 PM, Matt Ford <m...@dancingfrog.co.uk> wrote: >>>> I started by putting brackets in >>>> >>>> ([1,2] >>= \n -> [3,4]) >>= \m -> return (n,m) >>>> >>>> This immediately fails when evaluated: I expect it's something to do >>>> with the n value now not being seen by the final return. >>> >>> You're bracketing from the wrong end, which your intuition about n's >>> visibility hints at. Try this as your first set of parens: >>> >>> [1,2] >>= (\n -> [3,4] >>= \m -> return (n,m)) >>> >>> --Rogan >>> >>>> >>>> It seems to me that the return function is doing something more than >>>> it's definition (return x = [x]). >>>> >>>> If ignore the error introduced by the brackets I have and continue to >>>> simplify I get. >>>> >>>> [3,4,3,4] >>= \m -> return (n,m) >>>> >>>> Now this obviously won't work as there is no 'n' value. So what's >>>> happening here? Return seems to be doing way more work than lifting the >>>> result to a list, how does Haskell know to do this? Why's it not in the >>>> function definition? Are lists somehow a special case? >>>> >>>> Any pointers appreciated. >>>> >>>> Cheers, >>>> >>>> -- >>>> Matt >>>> >>>> _______________________________________________ >>>> Haskell-Cafe mailing list >>>> Haskell-Cafe@haskell.org >>>> http://www.haskell.org/mailman/listinfo/haskell-cafe >
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