这是Man or boy测试方案代码:
(define (A k x1 x2 x3 x4 x5)
(define (B)
(set! k (- k 1))
(A k B x1 x2 x3 x4))
(if (<= k 0)
(+ (x4) (x5))
(B)))
为了简化评估过程,我将其重写为:
(define (A k x1 x2)
(define (B)
(set! k (+ k -1))
(A k B x1))
(if (> 1 k)
(x2)
(B)))
我无法理解为什么(A 2 (lambda () 1) (lambda () -1))
会返回1。
任何人都可以解释Scheme解释器如何逐步评估此表达式。如果您可以附加环境图表,那就更好了:)
答案 0 :(得分:8)
The question is very subtle, and in the first moment I thought that the call would cause an infinte loop. But the real affair is the following:
Let's start calling F1 and F2 the two functions passed to A the first time, that is
F1 = (lambda() 1)
F2 = (lambda() -1)
So, after the first call of (A 2 F1 F2)
, A
establishes the following environment, that we will name E1
:
The test is now false, so A
calls B1
. B1
first decrements k
in E1
, then calls again A
, passing 1, itself, and x1
, which is F1
. So this is the call with parameters substituted with their values: (A 1 B1 F1)
. And the new environment established by this call (E2
) is shown in the following picture:
The test is still false, so A
calls B2
, which first modifies k
in E2
, then calls A
with 0, itself, and x1
(which now is B1
). So the call is (A 0 B2 B1)
, and the new set of environments is now:
The test is now true, so A
call x2
, which is B1
. Now B1
modifies k
in its environment (which is E1
), and then calls A
with 0, itself, and the value of x1
, which, in E1
, is F1
. So the call is (A 0 B1 F1)
and the environment established by this call is depicted in the next figure:
And finally, after checking that the test is true, A
calls x2
, that is F1
, which returns 1. At last!
答案 1 :(得分:2)
我建议在DrRacket的Stepper中检查程序。 Stepper允许您检查计算中的各个步骤。 你甚至可以回去。
唯一的问题是程序原样不能在步进器中运行。 因此我做了一些细微的修改,我希望它足够接近。
(define (A k x1 x2)
(local [(define (B _) (A (- k 1) B x1))]
(if (< k 2) (x2 '_) (B '_))))
(A 2 (lambda (_) 1) (lambda (_) -1))
该计划将在正常计划中:
(define (A k x1 x2)
(define (B _) (A (- k 1) B x1))
(if (< k 2) (x2 '_) (B '_)))
但教学语言要求您使用local
来进行本地定义。
此外,我已将参数_
添加到B
。它永远不会被使用。教学语言至少需要一个参数,所以我添加一个参数,然后忽略它。
以下是如何计算一步的屏幕截图: