回复： \a^b

Thanks. I'm doing \(X:Y)^p(X,Y) to try a lambda call with parameter pattern matching, not to try a binary lambda call. :-)
  

 

 ------------------ 原始邮件 ------------------
  发件人: "Ulrich Neumerkel";<[email protected]>;
 发送时间: 2013年12月12日(星期四) 凌晨1:06
 收件人: "Lumj"<[email protected]>; "SWI-Prolog"<[email protected]>; 
 
 主题: Re: [SWIPL] \a^b

 

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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