At 02:26 29/05/2007, you wrote:

PR Stanley wrote:

...

...

Hi

...

What is the C equivalent of the inequality operator in Haskell?

/=

You can answer these sorts of questions yourself using Hoogle: And what makes you think I haven't tried Google already? Unlike you I only write to the list when I have a legitimate reason to do so.

The respondent said HHHoogle, not Google, and rightly so. Did you take a look at the given link? Here it is again http://www.haskell.org/hoogle/?q=a%20-%3E%20a%20-%3E%20Bool

Hold onto that jerky knee! It's a sort of natural reaction to self-righteous jerks. Of course, a simple reply to the initial question would have been far too convenient.

At 02:27 29/05/2007, you wrote:

PR Stanley wrote:

...

...

Hi

...

What is the C equivalent of the inequality operator in Haskell?

/=

You can answer these sorts of questions yourself using Hoogle: And what makes you think I haven't tried Google already? Unlike you I only write to the list when I have a legitimate reason to do so.

Apparently, unlike him, you don't read the emails. He answered your question, gave you one way of answering all such future question (note Hoogle /= Google), and finally, there are certainly other places you could look before writing to the list that would answer your question, e.g. the Report. and you can mind your own damn business. What is this, the Haskell Cafe Gang?

...

...

Hi

...

What is the C equivalent of the inequality operator in Haskell?

/=

You can answer these sorts of questions yourself using Hoogle:

And what makes you think I haven't tried Google already? Unlike you I only write to the list when I have a legitimate reason to do so.

Just in case there was some sort of miscommunication, the actual answer to your question is (/=) :: a -> a -> Bool, as Neil said.

The /= wasn't some sort of frowny, quit-wasting-my-time emoticon. _______________________________________________ Thank you, Dan. Regards Paul

On Tue, May 29, 2007 at 11:20:27AM +0100, David House wrote:

Almost, (/=) :: Eq a => a -> a.

Well, not quite :) You forgot "-> Bool" at the end :)

(Just for completeness.)

Exactly :) -- Antti-Juhani Kaijanaho, Jyväskylä http://antti-juhani.kaijanaho.fi/newblog/

On Tue, May 29, 2007 at 11:20:27AM +0100, David House wrote:

On 29/05/07, Daniel McAllansmith wrote:

...

Just in case there was some sort of miscommunication, the actual answer to your question is (/=) :: a -> a -> Bool, as Neil said.

Almost, (/=) :: Eq a => a -> a.

Almost again! (/=) :: Eq a => a -> a -> Bool Best regards Tomek