op(box(x)) --> box, but is(op(box(x)) = box) --> false.
Subject: op(box(x)) --> box, but is(op(box(x)) = box) --> false.
From: Barton Willis
Date: Sat, 5 Nov 2011 19:56:13 -0500
It's a bug:
(%i13) ?print(op(box(x)));
MBOX
(%o13) box
A workaround:
(%i14) :lisp(defun $boxp (x) (and (consp x) (consp (car x)) (eq 'mbox (caar x))))
$BOXP
(%i14) boxp(box(x));
(%o14) true
--Barton
-----maxima-bounces at math.utexas.edu wrote: -----
To: maxima at math.utexas.edu
From: David R Stoutemyer
Sent by: maxima-bounces at math.utexas.edu
Date: 11/05/2011 07:10PM
Subject: op(box(x)) --> box, but is(op(box(x)) = box) --> false.
op(box(x)) --> box, but is(op(box(x)) = box) --> false.
?
Also is(op(box(x)) = 'box) --> false,
and is(op(box(x)) = "box") --> false.
?
How can I programmatically detect that op(expression) is box?
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