Re: Parameters

[Sebastian Tennant]

Quoth address@hidden (Ludovic Courtès):
> Sebastian Tennant <address@hidden> writes:
>> This simple procedure behaves as expected, i.e., it provides the sum of
>> a list of numbers:
>>
>>  (define add
>>    (lambda (l)
>>      (if (null? l)
>>          0
>>        (+ (add (cdr l)) (car l)))))
>>
>> whereas this procedure results in a stack overflow:
>>
>>  (define add
>>    (lambda l
>>      (if (null? l)
>>          0
>>        (+ (add (cdr l)) (car l)))))
>>
>> the only difference being the designation of the formal parameter of the
>> anonymous procedure; l or (l).
>
> When creating a procedure with `(lambda args body...)', then, no matter
> how it is invoked, ARGS will always be bound to the *list* of arguments
> that were passed to the procedure.  Consequently, such procedures can be
> passed any numbers of arguments.
>
> Conversely, `(lambda (x) body...)' is a one-argument procedure.  When it
> is invoked, X is bound to its argument.
>
> Thus, in your case, the first `add' would be used with a single
> argument, e.g., `(add '(1 2 3 4))', while the second would be used with
> an indefinite number of arguments, e.g., `(add 1 2 3 4)'.

Understood... up to a point.  However, I don't understand why calling
the first procedure like this:

 (add '(1 2 3 4))

and calling the second procedure like this:

 (add 1 2 3 4)

doesn't result in 'l' having the same value; '(1 2 3 4), in each case?

To put it another way, I would expect both procedures to work provided
they are called with (a) suitable argument(s).

Why does the second procedure fail regardless of how it is called?

> Note that none of your procedures is tail-recursive, so they consume
> stack space proportional to the length of L, which can lead to a stack
> overflow for large lists.  A tail-recursive definition would be:
>
>   (define add
>     (lambda (l)
>       (let loop ((l l)
>                  (result 0))
>         (if (null? l)
>             result
>             (loop (cdr l) (+ result (car l)))))))

Noted.

Sebastian