#1=(lambda () (car #1#)) is turned into (lambda nil #1=(car (lambda nil #1#))) which works the same, but has two more elements, am i correct? why does this happen?
Name:
Anonymous2014-10-19 2:46
or maybe it doesnt contain two more elements, but it does two passes to detect circularity, which results in elements being printed twice?
Name:
Anonymous2014-10-19 2:51
nope, its really two more elements :(
Name:
Anonymous2014-10-19 5:35
elisp
This shit shouldn't even exist anymore. There is Common Lisp, there is Scheme, there is Clojure. No need for a separate dialect just for the sake of emax.
>>9 touch this then, faggot. you like it raw i bet.
Name:
Anonymous2014-10-19 15:28
>>10 Reality check: you have no dubs. Prepare your anus!
Name:
Anonymous2014-10-20 4:22
i thought it was not possible, but you can evaluate circular code in elisp: (progn (setq a 10) #1=(print (if (>= (setq a (1- a)) 0) (progn (insert "fuck ") #1#) (insert "you ")))) fuck fuck fuck fuck fuck fuck fuck fuck fuck fuck you nil
nil
nil
nil
nil
nil
nil
nil
nil
nil
nil nil
Name:
Anonymous2014-10-20 4:24
forgot code tag (progn (setq a 10) #1=(print (if (>= (setq a (1- a)) 0) (progn (insert "fuck ") #1#) (insert "you ")))) fuck fuck fuck fuck fuck fuck fuck fuck fuck fuck you nil