为什么这些代码块不能全部工作？

Question

代码 #1 此代码有效

 (define *o
  (lambda (x y z)
    (if (= y 0)                                       
        z
        (*o x (sub1 y) (+ z x)))))

(*o 7 3 0)

code#2 此代码不起作用

(define *o (x y z)
    (cond
      ((0? y) z) 
      (else *o x (sub1 y) (+ z x))))                     

(print (*o 7 3 0))
(define *o
  (lambda (x y z)
    (cond
      ((0? y) z) 
      (else *o x (sub1 y) (+ z x))))) 

(*o 7 3 0)

code#3 此代码不起作用

(define *o (x y z)
    (cond
      ((0? y) z) 
      (else *o x (sub1 y) (+ z x))))                     

(print (*o 7 3 0))

#2 尝试使用

(cond

而不是

(if

#3 尝试重新表述代码，省略

(lambda

Answer 1

我将把你的第二个例子重写为

if

，这样你就可以看到做了什么：

(define *o (x y z)
  (if (0? y)         ; I'm guessing this is like (zero? y)
      z              ; consequent is to return z
      (begin         ; the alternative consiste of 4 expressions
        *o           ; evaluates to a procedure, then thrown away
        x            ; evaluates to a number, then thrown away
        (sub1 y)     ; evaluates to a number, then thrown away 
        (+ z x))))   ; tail expression evaluates to a number and returned 

(*0 1 2 3) ; ==> 4, due to (+ z x)
(*0 1 0 3) ; ==> 3, due to z

其意图可能是调用

*0

，但是您需要使用括号来应用它：

(define *o (x y z)
  (if (0? y)                    ; I'm guessing this is like (zero? y)
      z                         ; consequent is to return z
      (*o x (sub1 y) (+ z x)))) ; tail recursive call

条件：

(define *o (x y z)
    (cond
      ((0? y) z) 
      (else (*o x (sub1 y) (+ z x)))))