CODEWITHSUNDEEP
scheme
Tom
Tom

Reputation: 673

scheme - function with let

I saw that a similar question was asked before, but I don't understand another issue here.

Here are two functions:

(define x 2)
(define a 2)

(define goo
  (lambda (x)
    (display x)
    (lambda (y) (/ x y))))

(define foo
  (let ((f (goo a)))
    (lambda (x)
      (if (= x 0)
          x
          (f x)))))

What is the return value of (foo (foo 0))? What will print out to the screen?

As I understand it, when I run (foo 0) in the beginning, 2 will print out (we will enter the function goo), and the return value will be 0. Then, we will enter the function foo again with (foo (foo 0)) => (foo 0). We again enter the function goo and 2 will print out. But when I run it, 2 is printed just once. I think I'm missing a critical issue about let and its connection to lambda.

Upvotes: 1

Views: 2992

Answers (1)

Doug Currie
Doug Currie

Reputation: 41220

The let inside the definition of foo, and therefore the application of goo to a, is evaluated when foo is defined, not when foo is evaluated.

Look at it this way: what is the value of foo? It is the lambda expression. The binding of f is closed over by foo, it is not "redone" every time foo is evaluated.

Edit: here's an example without lambdas

> (let ((x (sqrt 2))) (* x 3))
4.24264068711929
> (define bar (let ((x (sqrt 2))) (* x 3)))
> bar
4.24264068711929
>

When you evaluate bar you are not calling sqrt again. bar is defined as the body of the let, in this case a number that is the result if an expression.

In your example the body of the let is a lambda expression. But just like my example, the let binding is not re-executed.

Upvotes: 2

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