On Tue, 18 Dec 2007 11:56:36 +0200, Jon Fairbairn <[EMAIL PROTECTED]> wrote:

"Cristian Baboi" <[EMAIL PROTECTED]> writes:


- the syntax for a block

Not sure what you mean by "block".

do a <- [1..10]
   b <- [3,4]
   return (a,b)

is an expression... you can write that same expression as
do {a <- [1..10]; b <- [3,4]; return (a,b)} too.


I mean anything that you can put between "{" "}", and between ";"


- the adhoc syntax rules (how to distinguish among a tuple
and a  pharanthesized expression

a tuple has commas in it. I'll grant that (x) not being a
1-tuple is a little ad-hoc, but there really is very little
ad-hockery in Haskell (and a 1-tuple behaves very much like
a plain value after all).

Is this ([1 ,2 ,3 ,4]) a tuple or what ?
It has commas in it!

- the fact that lambda expressions are not the same thing
as "algebraic data" values

It might help to know why you think they might be the same;
the syntax is different and the name is different...

Ah, just a thought, nothing more.
Lambda expressions are values, which is just data, after all.
Even C can apply a function variable to an argument (function pointers).


- what guarantees are made by the LANGUAGE that an IO action
(such as  do  putStrLn "Hello world" ) is not performed
twice

As has been pointed out, «do putStrLn "Hello world"» is an
expression that you can bind to a variable and use as many
times as you like. Incidentally, it means the same as
«putStrLn "Hello World"»; do connects a sequence of bindings
and expressions, so you don't need it if there's nothing to
be connected to.

Yes, but that was not the question.
What make you so sure it will be printed the exact number of times you intended ?

- the lambda expressions can be written (input) but cannot
be printed  (output)

This is a fundamental property of the language.  A lambda
expression is programme and at runtime the system doesn't
know one lambda expression from another (all it can do with
one is apply it to something).

Even C can apply a function variable to an argument (function pointers).
What make Haskell different beside the "lazy evaluation" and "mutable variables" things ?

The biggest problem for me, so far, is the last one.

I can't see how your example illustrates that, I'm afraid.

In a very strange way. Nevermind.

Here is some strange example:

What I don't understand is why I'm forced to use guards like
x==aa in cc,  when aa is clearly bounded (is 7) and why in
function h, the bounded u and  v become free variables in
the case expression.

I would have liked the language design to have permitted
case to pattern match against variables too, but the
question is, what would the syntax be?  There was a fair bit
of discussion about this when the language was designed (and
since), but no-one could come up with a good way of doing
it. One aspect of it is this: we want

f 0 = 42
f x = 3*x

to work, and we want all function definitions to be
translated into the core language in the same way,
so you get
f = \a -> case a of
            0 -> 42
            x -> 3*x

and given that, you can't have a variable on the LHS of ->
do anything other than get bound to the value of the
expression in the case (a in the example). It's not just a
the top level, either:

f Nothing = 0
f (Just n) = n+1

just means
f = \v -> case v of
            Nothing -> 0
            Just n -> n+1

so you can't have variables inside constructors do anything
but get bound at that point.




Thank you very much!


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