Can somebody explain ?

[Joel Foutelet]

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Hello

Can somedy explain that :

Consider the facts

to('->').
to('<-').
to('!').
to('a').

and the answers of the request to(A) :

A = (->) ;

 

A = <- ;

 

A = ! ;

 

A = a ;

Why -> is treated differently ?

 

Thank you

 

Joël

[Jean-Marc Vanel-2]

Cher Joël

-> is *not* treated differently, it is printed differently.
The reason is that is an operator visible by default.


2008/5/20 Joel Foutelet <joel.foutelet@...>:

> Hello
> Can somedy explain that :
> Consider the facts
> to('->').
> to('<-').
> to('!').
> to('a').
> and the answers of the request to(A) :
> A = (->) ;
>
> A = <- ;
>
> A = ! ;
>
> A = a ;
> Why -> is treated differently ?
>
> Thank you
>
> Joël

--
Jean-Marc Vanel
http://jmvanel.free.fr/
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SWI-Prolog mailing list
SWI-Prolog@...
https://mailbox.iai.uni-bonn.de/mailman/listinfo.cgi/swi-prolog

[Joel Foutelet]

OK thank you :-)

I noticed it was the same for =< , ==, >= ...

Joël
----- Original Message -----
From: "Jean-Marc Vanel" <jeanmarc.vanel@...>
To: "Joel Foutelet" <joel.foutelet@...>
Cc: <swi-prolog@...>
Sent: Wednesday, May 21, 2008 1:57 PM
Subject: Re: [SWIPL] Can somebody explain ?


Cher Joël

-> is *not* treated differently, it is printed differently.
The reason is that is an operator visible by default.


2008/5/20 Joel Foutelet <joel.foutelet@...>:

> Hello
> Can somedy explain that :
> Consider the facts
> to('->').
> to('<-').
> to('!').
> to('a').
> and the answers of the request to(A) :
> A = (->) ;
>
> A = <- ;
>
> A = ! ;
>
> A = a ;
> Why -> is treated differently ?
>
> Thank you
>
> Joël

--
Jean-Marc Vanel
http://jmvanel.free.fr/
_______________________________________________
SWI-Prolog mailing list
SWI-Prolog@...
https://mailbox.iai.uni-bonn.de/mailman/listinfo.cgi/swi-prolog

_______________________________________________
SWI-Prolog mailing list
SWI-Prolog@...
https://mailbox.iai.uni-bonn.de/mailman/listinfo.cgi/swi-prolog