Thanks for all the responses. They were indeed very helpful.
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Andrew J Bromage writes:
| G'day all.
|
| On Wed, Jun 05, 2002 at 10:35:52PM -0500, Jon Cast wrote:
|
| > > One general rule of strongly-typed programming is: A program is type
| > > correct if it is accepted by my favourite type checker. A corollary
| > > is that what you call a type, I
G'day all.
On Wed, Jun 05, 2002 at 10:35:52PM -0500, Jon Cast wrote:
> > One general rule of strongly-typed programming is: A program is type
> > correct if it is accepted by my favourite type checker. A corollary
> > is that what you call a type, I reserve the right to call a
> > precondition.
Andrew J Bromage <[EMAIL PROTECTED]> wrote:
> G'day all.
> On Wed, Jun 05, 2002 at 08:20:03PM -0500, Jon Cast wrote:
> > I think you're confused about what the type declarations mean.
> > When you say
> > > sqrt :: Float -> Float
> > you're promising to operate over /all/ Floats.
> That wou
G'day all.
On Wed, Jun 05, 2002 at 08:20:03PM -0500, Jon Cast wrote:
> I think you're confused about what the type declarations mean. When
> you say
>
> > sqrt :: Float -> Float
>
> you're promising to operate over /all/ Floats.
That would be true of Haskell functions were constrained to be
Cagdas Ozgenc <[EMAIL PROTECTED]> wrote:
> For example all functions with Int -> Int are type equivalent,
> because structural equivalency is used. Most of the time the
> functions are not compatible because they have different pre/post
> conditions, even though they seem to have the same input o
I remain puzzled by the thrust of your argument however one bit did
make sense:
> Also a function working over (Int,Int) will do so even if the
> numbers are totally irrelevant to that function. They maybe the
> number of (apples,oranges) or number of (books,authors).
It sounds like you want di
> For example all functions with Int -> Int are type equivalent
> However,
>
> data D a b = MkD a b
All objects D Int Int are type equivalent. I'm not sure what
your question means, otherwise.
If you define
data Function a b = F (a -> b)
apply:: Function a b -> a -> b
apply (F f) a = f a
y
For example all functions with Int -> Int are type equivalent, because
structural equivalency is used. Most of the time the functions are not
compatible because they have different pre/post conditions, even though they
seem to have the same input output types. I rather make my functions an
instanc
Cagdas Ozgenc writes:
| Greetings.
|
| I know 2 special type constructors(there might be other that I do
| not know yet) -> and ( , ) where structural type equivalency is
| enforced and we can also create new types with an algebric type
| constructor notation where name equivalency is enfor
At 20:11 05/06/2002 +0200, Hannah Schroeter wrote:
>Why not combine filter and length appropriately?
Pass -- I'm also a newbie, and haven't got as far as filter.
*Looks up filter*
Well, there's his homework done :-)
Hello!
On Wed, Jun 05, 2002 at 07:04:28PM +0100, Andy Fugard wrote:
> >= Original Message From "xoo" <[EMAIL PROTECTED]> =
> >hi.. i was just wondering if some body could give a simple equation for the
> following situation.other than recursion plz..
> >occurrences :: Eq a => a -> [a] -
Probably is a homework question, so HINT:
you can do it using 2 and only 2 prelude functions and no additional
function definitions of your own.
--
Hal Daume III
"Computer science is no more about computers| [EMAIL PROTECTED]
than astronomy is about telescopes." -Dijkstra | www.isi.edu/~
>= Original Message From "xoo" <[EMAIL PROTECTED]> =
>hi.. i was just wondering if some body could give a simple equation for the
following situation.other than recursion plz..
>
>occurrences :: Eq a => a -> [a] -> [a]
>--occurrences xs ys returns the number of times that xs occurs in ys
Greetings.
I know 2 special type constructors(there might be
other that I do not know yet) -> and ( , ) where structural type
equivalency is enforced and we can also create new types with an algebric type
constructor notation where name equivalency is enforced.
What is the rationale? I m
hi.. i was just wondering if some body could give a
simple equation for the following situation.other than recursion
plz..
occurrences :: Eq a => a -> [a] ->
[a]
--occurrences xs ys returns the number of times
that xs occurs in ys
thanks
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Sertifikam.Com Bilgisayar Eðitim MerkeziMicrosoft Certified Systems EngineerMicrosoft Certified Systems AdministratorMicrosoft Certified Database AdministratorMicrosoft Office User Web Designer Super Fiyatlar ! Ve http://www.sertifikam.com adresinden KAYIT olanlara %10 indirim
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