[Haskell-cafe] Syntax of 'do'

[Maurí­cio]

Hi,

http://haskell.org/haskellwiki/Keywords says that:

-
[do is a] syntactic sugar for use with monadic
expressions. For example:

 do { x ; result - y ; foo result }

is shorthand for:

 x  y = \result - foo result
-

I did some tests hiding Prelude. and Prelude.=
and applying  and = to non-monadic types, and
saw that 'do' would not apply to them. So, I would
like to add the following to that text:

-
as long as proper types apply:

x :: Prelude.Monad a
y :: Prelude.Monad b
foo :: b - Prelude.Monad c
-

Is that correct (Haskell and English)?

Thanks,
Maurício

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Re: [Haskell-cafe] Syntax of 'do'

[Miguel Mitrofanov]

-XNoImplicitPrelude ?

On 29 Aug 2008, at 17:41, Maurí cio wrote:


Hi,

http://haskell.org/haskellwiki/Keywords says that:

-
[do is a] syntactic sugar for use with monadic
expressions. For example:

do { x ; result - y ; foo result }

is shorthand for:

x  y = \result - foo result
-

I did some tests hiding Prelude. and Prelude.=
and applying  and = to non-monadic types, and
saw that 'do' would not apply to them. So, I would
like to add the following to that text:

-
as long as proper types apply:

x :: Prelude.Monad a
y :: Prelude.Monad b
foo :: b - Prelude.Monad c
-

Is that correct (Haskell and English)?

Thanks,
Maurício

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Re: [Haskell-cafe] Syntax of 'do'

[Bulat Ziganshin]

Hello Mauri­cio,

Friday, August 29, 2008, 5:41:41 PM, you wrote:

afaik, this shorthand isn't exact. actually there is more code dealing
with fails and this code use Monad operations


 Hi,

 http://haskell.org/haskellwiki/Keywords says that:

 -
 [do is a] syntactic sugar for use with monadic
 expressions. For example:

   do { x ; result - y ; foo result }

 is shorthand for:

   x  y = \result - foo result
 -

 I did some tests hiding Prelude. and Prelude.=
and applying  and = to non-monadic types, and
 saw that 'do' would not apply to them. So, I would
 like to add the following to that text:

 -
 as long as proper types apply:

 x :: Prelude.Monad a
 y :: Prelude.Monad b
foo :: b - Prelude.Monad c
 -

 Is that correct (Haskell and English)?

 Thanks,
 Mauricio

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 Haskell-Cafe mailing list
 Haskell-Cafe@haskell.org
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-- 
Best regards,
 Bulatmailto:[EMAIL PROTECTED]

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Re: [Haskell-cafe] Syntax of 'do'

[David House]

2008/8/29 Maurí­cio [EMAIL PROTECTED]:
 x :: Prelude.Monad a
 y :: Prelude.Monad b
 foo :: b - Prelude.Monad c

Monad is not a type, it is a type class, so you probably mean:

x :: Monad m = m a
y :: Monad m = m b
foo :: Monad m = b - m c

With the further understanding that all three `m's must be the same.

-- 
-David
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Re: [Haskell-cafe] Syntax of 'do'

[Philip Weaver]

On Fri, Aug 29, 2008 at 6:41 AM, Maurí­cio [EMAIL PROTECTED] wrote:

 Hi,

 http://haskell.org/haskellwiki/Keywords says that:

 -
 [do is a] syntactic sugar for use with monadic
 expressions. For example:

  do { x ; result - y ; foo result }

 is shorthand for:

  x  y = \result - foo result
 -

 I did some tests hiding Prelude. and Prelude.=
 and applying  and = to non-monadic types, and
 saw that 'do' would not apply to them. So, I would
 like to add the following to that text:


It sounds like you tried to redefine () and (=) and make 'do' use the
new definitions.  This is not possible, regardless of what types you give
() and (=).

If you want to define () and (=), do so for a particular instance of
Monad.


 -
 as long as proper types apply:

 x :: Prelude.Monad a
 y :: Prelude.Monad b
 foo :: b - Prelude.Monad c
 -

 Is that correct (Haskell and English)?

 Thanks,
 Maurício

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 Haskell-Cafe@haskell.org
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Re: [Haskell-cafe] Syntax of 'do'

[David House]

2008/8/29 Philip Weaver [EMAIL PROTECTED]:
 It sounds like you tried to redefine () and (=) and make 'do' use the
 new definitions.  This is not possible, regardless of what types you give
 () and (=).

Watch out for rebindable syntax:
http://www.haskell.org/ghc/docs/latest/html/users_guide/syntax-extns.html#rebindable-syntax

At first reading, I thought that -XNoImplicitPrelude was required to
turn this on. But now I'm not sure: it seems that if you hide
Prelude.= and Prelude.return, that ought to be enough to make do
notation work with your alternative definitions. I'm not at home, so I
can't try this right now.

-- 
-David
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Re: [Haskell-cafe] Syntax of 'do'

[Philip Weaver]

On Fri, Aug 29, 2008 at 8:50 AM, David House [EMAIL PROTECTED] wrote:

 2008/8/29 Philip Weaver [EMAIL PROTECTED]:
  It sounds like you tried to redefine () and (=) and make 'do' use the
  new definitions.  This is not possible, regardless of what types you give
  () and (=).

 Watch out for rebindable syntax:

 http://www.haskell.org/ghc/docs/latest/html/users_guide/syntax-extns.html#rebindable-syntax



Oh, I had no idea!  Thanks :).



 At first reading, I thought that -XNoImplicitPrelude was required to
 turn this on. But now I'm not sure: it seems that if you hide
 Prelude.= and Prelude.return, that ought to be enough to make do
 notation work with your alternative definitions. I'm not at home, so I
 can't try this right now.



 --
 -David

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