Re: \a^b

Ulrich Neumerkel <[email protected]>
Newsgroups gmane.comp.ai.prolog.swi
Message-ID <[email protected]>
Lumj :
>In 6.4.1, I get following:
>?-\a^b=..L.
>L=[\,a^b].
>?-\(a)^b=..L.
>L=[^,\a,b].
>
>As \ and ^ both have a precedence of 200, i want to know if there's
> any theory in the result? Because I have a tendency to think that the
> parentheses do not affect the association of terms outside them.

\(a)^b is like (\(a))^b because no space before the opening bracket
enforces functional notation, disabling the operators.


The precedence specifier is what is of relevance here:

:- op(200,xfy,^).
:- op(200, fy,\).

>In fact this is preventing using lambda like this:
>call(\(X:Y)^p(X,Y),x,y).

Why the : ? 

call(\X^Y^p(X,Y),x,y) does this with

http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/lambda.pl
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