On Fri, Jul 19, 2013 at 3:58 PM, Matt Ford 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 wrote:

On Fri, Jul 19, 2013 at 3:23 PM, Matt Ford 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

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I thought >>= was left associative? It seems to be in the examples from Learn You A Haskell.

It is. But lambdas are parsed using the "maximal munch" rule, so they extend *as far to the right as possible*. So \x -> x * 2 + 1 would be parsed as \x -> (x * 2 + 1) -- right not (\x -> x) * 2 + 1 -- wrong which is obviously incorrect. I believe C uses a similar rule for funny expressions like `x+++y` (using maximal munch: `(x++) + y`).

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))

Any thoughts?

Matt

On 19 Jul 2013, at 23:35, Rogan Creswick wrote:

On Fri, Jul 19, 2013 at 3:23 PM, Matt Ford 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

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-- Chris Wong, fixpoint conjurer e: lambda.fairy@gmail.com w: http://lfairy.github.io/

For the sake of approaching this in yet another way, it can also be helpful to substitute the definitions of bind and return in your expression. If we start with the definitions: instance Monad [] where xs >>= f = concat (map f xs) return x = [x] Then we can make the following transformations: [1,2] >>= \n -> [3,4] >>= \m -> return (n,m) [1,2] >>= \n -> [3,4] >>= \m -> [(n, m)] [1,2] >>= \n -> concat (map (\m -> [(n, m)]) [3,4]) concat (map (\n -> concat (map (\m -> [(n, m)]) [3,4])) [1,2]) Or perhaps more simply: concatMap (\n -> concatMap (\m -> [(n, m)]) [3,4]) [1,2] All of which are valid expressions and produce the same value. Depending on your learning style this might not be as helpful as the other approaches, but it does take a lot of the mystery out of >>= and return. On Sat, Jul 20, 2013 at 1:08 AM, Alberto G. Corona wrote:

Matt

It is not return, but the bind the one that does the miracle of multiplication. 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 lists generated

2013/7/20 Matt Ford

...

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.

Cheers,

-- Matt

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

-- Alberto.

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