How is set! defined in scheme?

Question

How would you implement your own set! function in Scheme? A set! function is a destructive procedure that changes a value that is defined taking into account th

Answer 1

As has been mentioned, set! is a primitive and can't be implemented as a procedure. To really understand how it works under the hood, I suggest you take a look at the inner workings of a Lisp interpreter. Here's a great one to start: the metacircular evaluator in SICP, in particular the section titled "Assignments and definitions". Here's an excerpt of the parts relevant for the question:

(define (eval exp env)
  (cond ...
        ((assignment? exp) (eval-assignment exp env))
        ...
        (else (error "Unknown expression type -- EVAL" exp))))

(define (assignment? exp)
  (tagged-list? exp 'set!))

(define (eval-assignment exp env)
  (set-variable-value! (assignment-variable exp)
                       (eval (assignment-value exp) env)
                       env)
  'ok)

(define (set-variable-value! var val env)
  (define (env-loop env)
    (define (scan vars vals)
      (cond ((null? vars)
             (env-loop (enclosing-environment env)))
            ((eq? var (car vars))
             (set-car! vals val))
            (else (scan (cdr vars) (cdr vals)))))
    (if (eq? env the-empty-environment)
        (error "Unbound variable -- SET!" var)
        (let ((frame (first-frame env)))
          (scan (frame-variables frame)
                (frame-values frame)))))
  (env-loop env))

In the end, a set! operation is just a mutation of a binding's value in the environment. Because a modification at this level is off-limits for a "normal" procedure, it has to be implemented as a special form.