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newLISP Fan Club

Forum => Anything else we might add? => Topic started by: ssqq on January 07, 2015, 06:10:19 AM

Title: *(quote fn) or (quote lambda)* error
Post by: ssqq on January 07, 2015, 06:10:19 AM
Following code would throw error:


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> 'fn
ERR: invalid lambda expression : "fnn"
> 'lambda
ERR: invalid lambda expression : "lambdan"
> (quote fn)
ERR: invalid lambda expression : " fn)n"
> (quote lambda)
ERR: invalid lambda expression : " lambda)n"


If it is a bug?
Title: Re: *(quote fn) or (quote lambda)* error
Post by: rickyboy on January 07, 2015, 09:48:48 AM
Lutz explained why this so, not too long ago here: http://www.newlispfanclub.alh.net/forum/viewtopic.php?f=5&t=4553&p=22482#p22482



In short, fn and lambda cannot be symbols in the sense you are using them.  If you are trying to build newLISP lambdas in your macro code, then follow Lutz's advice in the link above.
Title: Re: *(quote fn) or (quote lambda)* error
Post by: ssqq on January 07, 2015, 04:53:11 PM
Thanks, I use *lambda?* to check if expression starts with *fn* or *lambda*.
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(lambda? '(fn (x) (add x))) ;--> true
to get the symbol of *fn* or *lambda*, could use following code:


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(sym "fn") ; --> fn
(sym "lambda") ; --> lambda

then, could check it.


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> > (list 1 (sym "lambda") (sym "fn"))
(1 lambda fn)
> (= (sym "fn") (last (list 1 (sym "lambda") (sym "fn"))))
true
> (= (sym "lambda") (nth 1 (list 1 (sym "lambda") (sym "fn"))))
true
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