[Haskell-cafe] List Monads and non-determinism
Alberto G. Corona
agocorona at gmail.com
Sat Jul 20 10:08:46 CEST 2013
It is not return, but the bind the one that does the miracle of
By its definition for the list monad, it applies the second term once for
each element are in the first term.
So return is called many times. At the end, bind concat all the small
2013/7/20 Matt Ford <matt at dancingfrog.co.uk>
> Hi All,
> I thought I'd have a go at destructing
> [1,2] >>= \n -> [3,4] >>= \m -> return (n,m)
> which results in [(1,3)(1,4),(2,3),(2,4)]
> 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.
> 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.
> Haskell-Cafe mailing list
> Haskell-Cafe at haskell.org
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the Haskell-Cafe