问题
I've spent a day reading page 166's length≤1
in the book The Little Schemer; there's the following code:
(((lambda (mk-length)
(mk-length mk-length))
(lambda (mk-length)
(lambda (l)
(cond
((null? l) 0)
(else (add1
((mk-length eternity)
(cdr l))))))))
l)
where l
is (apples)
and eternity
is as follows:
(define eternity
(lambda (x)
(eternity x)))
Page 166 (4th ed.) states that:
When we apply
mk-length
once, we getlength≤1
And then
Could we do this more than once?
But I do not know how to do this for getting length≤2
?
回答1:
Suppose l
is (apples oranges)
, then it will evaluate like this (note that mk-length
is bound to the the (lambda (mk-length) ...)
function itself:
(cond ((null? l) 0)
(else (add1 ((mk-length eternity) (cdr l)))))
==>
(add1 ((mk-length eternity) '(oranges)))
==>
(add1 ((lambda (l) (cond ((null? l) 0
(else (add1 ((eternity eternity) (cdr l))))))))
==>
(add1 (add1 ((eternity eternity) '())))
So here, after two steps, eternity
ends up being applied, but what we want is for it to call mk-length
. So in the original function, if we replace eternity
by mk-length
, then the last step I wrote will contain (mk-length mk-length)
instead of (eternity eternity)
, allowing the computation to proceed.
来源:https://stackoverflow.com/questions/12564013/how-to-do-this-length%e2%89%a41-more-than-once