Hi Mark,
| I have just implemented the function "penultimax" which takes a list
| of positive integers and produces the "penultimate maximum", that is,
| the next biggest integer in the list after the maximum. Eg:
|
| penultimax [15,7,3,11,5] = 11
To your three implementations, let me add anoth
Hi,
I have just implemented the function "penultimax" which takes a list
of positive integers and produces the "penultimate maximum", that is,
the next biggest integer in the list after the maximum. Eg:
penultimax [15,7,3,11,5] = 11
One implementation is:
penultimax :: [Int] -> Int
penultimax
You caught my attention.
It would be nice if you write your own version from scratch to make all of us
profit of this.
Thank you.
Francis Girard
Le Conquet
France
Le 22 Novembre 2002 16:03, George Russell a écrit :
>
> I think in fact I'm going to give up on stealing other people's code and
>
--- Lu Mudong <[EMAIL PROTECTED]> wrote:
> Thanks a lot for you guys' help.
>
> I am very new to haskell and tried some methods you
> guys advised, doesn't
> seem to work, i think i didn't do it properly,
> here's my code and result,
> hope you can point out what's wrong. thanks!
>
Lots of theo
=
International Conference on
Geometric Modeling & Graphics
(Previously, Computer Aided Geometric Design Symposium)
(GMAG 2003)
in co-operation with IEEE Computer Society
JULY 16-18, 2003
London, UK.
=
Aims and Scope:
===
Computer-Aided (CA) tec
On Sun, 24 Nov 2002 20:42:31 +
Glynn Clements <[EMAIL PROTECTED]> wrote:
> Even if Haskell were strict, you still wouldn't be able to treat I/O
> operations as functions without discarding referential transparency.
Yes, but if haskell were strict, it wouldn't really need referential
transpar
Nick Name wrote:
> This approach is called "monads" and is needed because haskell is a lazy
> language, so order of evaluation is unspecified, while input/output
> usually needs a precise order of evaluation.
It is needed because Haskell is a functional language, where functions
are referentiall
Lu Mudong wrote:
> hi, all, I'm having problem convert IO String to string,
That's because you can't convert IO String to String.
> I read a file, and it's an IO String,
No.
Calling readFile doesn't read the file, and the value which readFile
returns isn't the result of reading the file. The
Hello,
several mails spoke about converting an IO String to a String. I have to point
out that such a conversion is not what happens when you use monadic I/O. In
fact, such a conversion is just not possible in Haskell 98.
Since Haskell is a pure language, evaluating a String expression mustn't
On Sun, 24 Nov 2002 09:05:17 -0900
"Lu Mudong" <[EMAIL PROTECTED]> wrote:
> Thanks a lot for you guys' help.
>
> I am very new to haskell and tried some methods you guys advised,
> doesn't seem to work, i think i didn't do it properly, here's my code
> and result, hope you can point out what's wr
Thanks a lot for you guys' help.
I am very new to haskell and tried some methods you guys advised, doesn't
seem to work, i think i didn't do it properly, here's my code and result,
hope you can point out what's wrong. thanks!
my code:
myReadFile :: IO String
myReadFile = readFile "d:/hugs98/in
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