I'm studying 'trace in @lib.l, and I've run into a question.
I don't understand why must 'eval be called on the second number on the
case that the first number is negative. I'm especially confused by giving 1
as an argument to 'eval. That's supposed to give you the context from which
to extract the value of @, but since its a number, it should never have an
@ to affect the result. If it was an expression, then I could understand,
but I've tried passing an expression as a second argument several times and
its always failed.
My question would be, why eval the second argument, and why pass a 1 to
eval when @ does not seem to be relevant in this context?
Here's the body of the 'trace function, as a refresher.
(de task (Key . Prg)
(nond
(Prg (del (assoc Key *Run) '*Run))
((num? Key) (quit "Bad Key" Key))
((assoc Key *Run)
(push '*Run
(conc
(make
(when (lt0 (link Key))
(link (+ (eval (++ Prg) 1))) ) ) #the line that
confuses me
(ifn (sym? (car Prg))
Prg
(cons
(cons 'job
(cons
(lit
(make
(while (atom (car Prg))
(link
(cons (++ Prg) (eval (++ Prg) 1)) )
) ) ) # the same pattern in another line.
Prg ) ) ) ) ) ) )
(NIL (quit "Key conflict" Key)) ) )
