SICP example doesn't work on Racket

Question

I am trying an example on Chapter 4 of SICP (part of writing the LISP interpreter)

(define (definition-value exp)
    (if (symbol? (cadr exp))
        (caddr exp)
        (make-lambda 
                    (cdadr exp) ; formal parameters
                    (cddr exp)  ; body
        )              
    )
) 


(define (make-lambda parameters body)
    (cons 'lambda (cons parameters body))
)

I Tested it, definition-value on '(define (double x) (+ x x))) should return a lambda function

( (definition-value '(define (double x) (+ x x))) 10)

Racket outputs

procedure application: expected procedure, given: (lambda (x) (+ x x)); arguments were: 10

Isn't "(lambda (x) (+ x x))" a procedure? Or it is a reference? If it is a reference, any way to "dereference" it?

Will Ness Will Ness · Accepted Answer · 2013-05-14T09:30:10

definition-value returns the value in the definition expression given to it as an argument:

(definition-value '(define x 42))  
=> 42

(definition-value '(define (qq x) (+ x y 42))) 
=> (make-lambda '(x) '((+ x y 42)))
=> '(lambda (x) (+ x y 42))

You can't call the quoted list as a function, as you do: ( '(lambda (x) (+ x y 42)) 10) is invalid. It is not a function, it is just an s-expression.

definition-value is part of an interpreter. This interpreter is the way to "dereference", i.e. interpret function definitions. Different interpreters can have different ways to interpret same function definitions, giving different semantics to the resulting languages.

Evaluation of expressions must be done in context - they appear inside certain lexical scope (area in code where a variable is visible), which gives rise to environments (also, this). In the example above, y is defined in some enclosing scope in the program being interpreted. Trying to interpret that expression in REPL by calling Racket's eval, what value would y have?

SICP example doesn't work on Racket

2 Answers