方案中的递归函数构成 [英] Recursive Function Composition in Scheme
问题描述
下面是我为创建一个程序返回给定函数列表中的函数组合的尝试。我陷入了僵局;我写的东西在纸上看起来很有意义,但是我没有看到我出错的地方,任何人都可以给出一些提示?
; (compose-all-rec fs) - >程序
; fs:listof过程
;返回fs中所有函数的函数组合:
;如果fs =(f0 f1 ... fN),结果为f0(f1(...(fN(x))...))
; (定义compose-all-rec(lambda(fs))
(if(empty?fs)empty
(lambda(fs)
(lambda(fs))
申请(第一个fs)(compose-all-rec(rest fs)))
))))
其中((compose-all-rec(list abs inc))-2)应该等于1
我会尝试一种不同的方法:
(define(compose-all-rec fs)
(define(apply-all fs x)
(if(empty?fs)
x
((first fs)(apply-all(rest fs)x))))
(λ(x)(apply-all fs x) ))
请注意,单个 lambda
需要在最后被返回,它在lambda中(它捕获 x
参数和 fs
列表),发生所有函数的实际应用 - 使用 apply-all
辅助程序。另请注意,(apply fx)
可以更简洁地表示为(fx)
。
如果允许更高阶的程序,更短的解决方案可以用 foldr
和一些语法糖用于返回咖喱功能:
(define((compose-all-rec fs)x)
(foldr(λ(fa)(fa))x fs))
无论哪种方式,建议的解决方案都按预期工作:
<$ p $ ((compose-all-rec(list abs inc))-2)
=> 1
Below is an attempt I've made to create a procedure that returns the function composition given a list of functions in scheme. I've reached an impasse; What I've written tried makes sense on paper but I don't see where I am going wrong, can anyone give some tips?
; (compose-all-rec fs) -> procedure
; fs: listof procedure
; return the function composition of all functions in fs:
; if fs = (f0 f1 ... fN), the result is f0(f1(...(fN(x))...))
; implement this procedure recursively
(define compose-all-rec (lambda (fs)
(if (empty? fs) empty
(lambda (fs)
(apply (first fs) (compose-all-rec (rest fs)))
))))
where ((compose-all-rec (list abs inc)) -2) should equal 1
I'd try a different approach:
(define (compose-all-rec fs)
(define (apply-all fs x)
(if (empty? fs)
x
((first fs) (apply-all (rest fs) x))))
(λ (x) (apply-all fs x)))
Notice that a single lambda
needs to be returned at the end, and it's inside that lambda (which captures the x
parameter and the fs
list) that happens the actual application of all the functions - using the apply-all
helper procedure. Also notice that (apply f x)
can be expressed more succinctly as (f x)
.
If higher-order procedures are allowed, an even shorter solution can be expressed in terms of foldr
and a bit of syntactic sugar for returning a curried function:
(define ((compose-all-rec fs) x)
(foldr (λ (f a) (f a)) x fs))
Either way the proposed solutions work as expected:
((compose-all-rec (list abs inc)) -2)
=> 1
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