I think 'curry' is what you're looking for. Your 'adder' example could be written as:
: (de adder (@N) (curry (@N) (X) (+ X @N))) -> adder : (adder 3) -> ((X) (+ X 3)) : (doc 'curry) # for more info :) Hope that helps, Erik On Feb 7, 2017 10:04 PM, "pd" <eukel...@gmail.com> wrote: > Hello, > > I wonder if there is any way to bind a free symbol in a lambda in order to > pass the lambda to a defined function (for example) > > What I want to do is something like this: > > (de myf (f l) (f l)) > > (let l 99 (myf '((x) (+ (car x) l)) (1 2))) > > I want it to return 100 but it fails with an error (1 2) number expected > > this is because free symbol l in lambda ((x) (+ (car x) l)) is not bind by > let as pretended because of quoting of lambda > > In other words, I think the problem is quoting avoids let binding as in: > > (setq f (let n 10 '((x) (+ x n)))) -> ((x) (+ x n)) > > but I want it to return -> ((x) (+ x 10)) > > or using the typicall example: > > (de adder (n) '((x) (+ x n))) -> ((n) '((x) (+ x n))) > > so (adder 1) should return ((x) (+ x 1)) but it returns ((x) (+ x n)) > > Is there any way to manage this? Something similar to "expand" in newlisp > will do the job: > > newlisp: > (define (badadder n) (lambda (x) (+ x n))) > (badadder 3) -> (lambda (x) (+ x n)) > (define (adder n) (expand (lambda (x) (+ x n)) 'n)) > (adder 3) -> (lambda (x) (+ x 3)) > > The fist example in newlisp will be: > > (define (myf f l) (f l)) > > (let ((l 99)) (myf (lambda (x) (+ (first x) l)) '(1 2))) > > which fails for same reason picolisp fails but in newlisp the solution is: > > (define (myf f l) (f l)) > > (let ((l 99)) (myf (expand (lambda (x) (+ (first x) l)) 'l) '(1 2))) -> > 100 > > > thanks > > >
