方案字符串追加?递归复制字符串 [英] scheme string-append? recursion to replicate a string

问题描述

设计一个名为 string-dup 的程序，它使用一个字符串 s 和一个数字 n 并返回一个字符串，该字符串是 s n 次的串联，每个 s 实例之间有空格，即，

Design a program called string-dup that consumes a String s and a Number n and returns a String that is the concatenation of s n times with spaces between each instance of s, i.e.,

(string-dup "a" 3) => "a a a"不使用复制，但我想我们可以使用字符串追加.

(string-dup "a" 3) => "a a a" Without using replicate but i guess we can use string-append.

到目前为止我得到了

(define (string-dup s n)
(cond 
  [(zero? n) ""]
  [else
   (cond
     [(= n 1 ) s]
     [(> n 1 )(string-dup (string-append s " ") (sub1 n))])]))

然而，这只允许我有一个字符串a".我知道你什么时候可以做列表案例

However, this only allows me to have one string "a". I know when in the list case you can do

(cons s (string-cp s (sub1 n))) 

但是在这种情况下如何申请?谢谢.

but how to apply in this case? Thanks.

推荐答案

string-dup 的递归调用需要是 string-append 的参数，而不是另一边.这是一个更正(和重构)的版本:

The recursive call to string-dup needs to be a parameter to string-append, not the other way round. Here's a corrected (and refactored) version:

(define (string-dup s n)
  (cond 
    [(< n 1) ""]
    [(= n 1) s]
    [else    (string-append s " " (string-dup s (sub1 n)))]))

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