回复： \a^b

Cool. So working with \ (X:Y)^p(X,Y) is just fine. Thanks for your quick response:-)
  

 

 ------------------ 原始邮件 ------------------
  发件人: "Jan Wielemaker";<[email protected]>;
 发送时间: 2013年12月11日(星期三) 晚上9:17
 收件人: "Lumj"<[email protected]>; "SWI-Prolog"<[email protected]>; 
 
 主题: Re: [SWIPL] \a^b

 

On 12/11/2013 01:56 PM, Lumj wrote:
> 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.
> In fact this is preventing using lambda like this:
> call(\(X:Y)^p(X,Y),x,y).

It is a beautyful example :-)  The trouble is that \(a) is read
name(arg1, ...) rather than a prefix operator. So, in
\a^b, both \ and ^ are operators and the precedence tells
the system how to interpret this.  In \(a)^b, \(a) is a
compound which has precedence 0.  That is also why you
have to write

\+ (a,b)

rather than

\+(a,b)

which reads as \+ with two arguments.

Cheers --- Jan


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