I'm working on a scheme evaluator in scheme. I need to implement let, I have parsed so that I have variable names, values to input and the body of the function. I need to return a lambda function using the parsed information and as such I have the following code:
(define (eval-let exp env)
((lambda (let-variables (let-bindings exp)) (let-body exp)) (let-exp (let-bindings exp))))
(let-variables (let-bindings exp)) evaluates to a list of variable names (ex: '(x y)), so I'm basically evaluating to this:
((lambda '(x y) (* x y)) '(2 3))
The scheme interpreter simple says: #%plain-lambda: not an identifier in: (let-bindings exp) which I'm guessing is because it wants a set of identifiers, not a list of values.
How do I turn my list of values into a set of identifiers?
To implement a let expression in your own interpreter first you have to transform it into a lambda application, something similar to this (procedure names should be self-explanatory):
(define (let->combination exp)
(let* ((bindings (let-bindings exp))
(body (let-body exp))
(exps (bindings-all-exps bindings))
(lambda (make-lambda (bindings-all-vars bindings) body)))
(make-application lambda exps)))
(define (make-lambda parameters body)
(list* 'lambda parameters body))
(define (make-application proc . args)
(cond ((null? args) (list proc))
((list? (car args)) (cons proc (car args)))
(else (cons proc args))))
And after the syntax transformation has been performed you can proceed to evaluate it:
(eval (let->combination exp) env)
What I'm trying to point out is that you shouldn't try to evaluate it directly. Also be careful, the code you're generating has a couple of incorrect quotes:
((lambda '(x y) (* x y)) '(2 3))
^ ^
here here
It should look like this instead:
((lambda (x y) (* x y)) 2 3)