CODEWITHSUNDEEP
recursionlispschemetail-recursion
Houdini
Houdini

Reputation: 3542

Beginner Scheme: Procedures that return themselves

This is an example from the book I am reading:

1    (define (length items)
2     (define (length-iter a count)
3       (if (null? a)
4           count
5           (length-iter (cdr a)(+ 1 count))))
6      (length-iter items 0))

What I am not understanding is how can length-iter know about count? The first time this procedure is called with a list, it will in turn define another procedure with two argumenets, I get that much. But how does it know that a is the list items? It hasnt reached line 6 yet, where items is passed to length-iter as the argument a. Somehow though it already knows this and is able to make the computation. Any help in clarifying this a bit is appreciated!

Upvotes: 1

Views: 125

Answers (1)

João Silva
João Silva

Reputation: 91349

There are two parts in the length function:

  1. Definition of the inner function length-iter;
  2. Invocation of the inner function length-iter.

In the invocation, i.e., line 6, you pass the original items list to the inner function as an argument. This is where the inner function gets called. Previously, you are just defining the function, not calling it:

(length-iter items 0)

Thus, items will be bound to a, and 0 to count. You can think of the inner function as a separate function:

(define (length-iter a count)
  (if (null? a)
      count
      (length-iter (cdr a)(+ 1 count))))

And then, think of your length function as if it just delegated all the work to the length-iter function:

(define (length items)
  (length-iter items 0))

That's what's being done in your function. The difference, is that the length-iter function is only known to length.

Upvotes: 4

Related Questions