On 07/20/2013 12:23 AM, Matt Ford wrote:
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 is not the same expression any more. See below for the correct
bracketing.
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.
...
[1,2] >>= (\n -> [3,4] >>= (\m -> return (n,m)))
~>*
((\n -> [3,4] >>= (\m -> return (n,m))) 1) ++ ((\n -> [3,4] >>= (\m ->
return (n,m))) 2)
~>*
([3,4] >>= (\m -> return (1,m))) ++ ([3,4] >>= (\m -> return (2,m)))
~>*
((\m -> return (1,m)) 3 ++ (\m -> return (1,m)) 4) ++ ((\m -> return
(2,m)) 3 ++ (\m -> return (2,m)) 4)
~>*
return (1,3) ++ return (1,4) ++ return (2,3) ++ return (2,4)
~>*
[(1,3)] ++ [(1,4)] ++ [(2,3)] ++ [(2,4)]
~>*
[(1,3),(1,4),(2,3),(2,4)]
Where the definition return x = [x] has been applied in the second-last
step.
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