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 wro
the tricky part then is to add support for other types.
another approach to existentially package type classes with the data type!
eg
data HasEq = forall a . HasEq ( Eq a => a)
or its siblinng
data HasEq a = Haseq (Eq a => a )
note this requires more planning in how you structure your program,
On 07/16/2013 04:57 PM, Michael Orlitzky wrote:
>
> This all works great, except that when there's 20 or so options, I
> duplicate a ton of code in the definition of OptionalCfg. Is there some
> pre-existing solution that will let me take a Cfg and create a new type
> with Cfg's fields wrapped in
On 07/20/2013 12:58 AM, 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.
...
Yes, >>= is left-associative. The associativity of >>= is not relevant
for your example because no two >>= operations actually o
> 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
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] >>= (
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))
Any thoughts?
Matt
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
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
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
On Fri, Jul 19, 2013 at 5:19 AM, Jose A. Lopes wrote:
> Hello,
>
> How to define equality for Data.Dynamic ?
Hi Jose,
You could try casting the values to different types that do have an
(==). You can treat the case where you have the types matching, but
didn't list that type beforehand different
Hi,
Am Freitag, den 19.07.2013, 11:19 +0200 schrieb John Blackbox:
> The question about generating the code was only to have a "debugging
> tool" - to see if the generated AST is good - I wanted to generate the
> Haskell code only to check if its correct, but normally I would not do
> it, because
Hi John,
> Alan - I do NOT want to generate Haskell code. I want only to generate AST
> and compile it.
> The question about generating the code was only to have a "debugging tool"
> - to see if the generated AST is good - I wanted to generate the Haskell
> code only to check if its correct, but
I accidentally didn't send that email to haskell-cafe, so I'm pasting it
here also:
Alan - I do NOT want to generate Haskell code. I want only to generate AST
and compile it.
The question about generating the code was only to have a "debugging tool"
- to see if the generated AST is good - I wanted
Thank you!
So, if I'm writing a compiler of custom language, which I want to generate
Haskell AST and further compile it with GHC, you prefer something like
haskell-src-exts
over pure GHC API?
2013/7/19 Antoine Latter
> The package haskell-src-exts is a lot less intimidating if all you are
> tr
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