CODEWITHSUNDEEP
lambdascheme
John K
John K

Reputation: 506

Is this the correct way to create a lambda from two lists in scheme?

I'm trying to create a lambda from parameters such that (apply (make-lambda '(a b) '(+ a b)) '(1 2)) returns 3

My first attempt (define (make-lambda params func) (lambda params func)) returns (+ a b)

This code works but it doesn't feel right to me: (define (make-lambda params func) (eval (list 'lambda params func)))

What is the correct way of doing this?

Upvotes: 1

Views: 86

Answers (2)

leppie
leppie

Reputation: 117350

Because I am bored, and if your scheme supports syntax-case :) 

(define-syntax make-lambda
  (lambda (x)
    (syntax-case x (quote)
      [(_ (quote pars) (quote body))
         #'(lambda pars body)])))

PS: Will probably work with syntax-rules too (see Óscar López's answer). So I am lazy too.

Example: http://eval.ironscheme.net/?id=186

Here is a syntax-rules version:

(define-syntax make-lambda
  (syntax-rules (quote)
    [(_ (quote pars) (quote body))
      (lambda pars body)]))

Example: http://eval.ironscheme.net/?id=187

Note: keep in mind, this is exactly what you asking for :D

Upvotes: 1

Óscar López
Óscar López

Reputation: 236150

You have to evaluate the list to use it as a procedure, otherwise is just a list of data which happens to have the symbol lambda as the first element. This syntax is a bit simpler, but is essentially the same:

(define (make-lambda params func)
  (eval `(lambda ,params ,func)))

This should work:

(apply (make-lambda '(a b) '(+ a b)) '(1 2))
=> 3

Alternatively, using a macro as suggested by Leppie:

(define-syntax make-lambda
  (syntax-rules ()
    ((_ args body)
     (lambda args body))))

Use it like this:

(apply (make-lambda (a b) (+ a b)) '(1 2))
=> 3

Upvotes: 3

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