Turning a list of list to list of string using syntax->string

Basically, I want '( (whatever1) (whatever2) (whatever3) ... ) ===> ( "(whatever1)" "(whatever2)" "(whatever3)" ), which is just add quotes outside of the list, 开发者_StackOverflow社区and keep the contents in the list unchanged. e.g.

'((define X ::int)
 (define b0 :: bool (=> T (and (= X X) (= 0 0)))))

will be turned into:

'("(define X ::int)"
 "(define b0 :: bool (=> T (and (= X X) (= 0 0))))")

However, the following code I am using eliminate all spaces!

#lang racket
(require syntax/to-string)
(define lst-sub '((define x :: int) (=> T (and (= X X) (= 0 0)))))
(pretty-write (map (λ (x) (string-append "(" (syntax->string (datum->syntax #f x)) ")")) lst-sub))  

which returns

("(definex::int)" "(=>T(and(=XX)(=00)))")

So the question is: there is no spaces anymore! How can I get around this??

#lang racket
(define lst-sub '((define x :: int) (=> T (and (= X X) (= 0 0)))))
(pretty-write (map (λ (x) (format "~s" x)) lst-sub))

Alright. I don't take the "easy" route I thought. and worked out as follows, which ends up with more lines of code :(

(define (toString-with-space data)
  (match data
    [(? symbol?) (string-append (symbol->string data) " ")]
    [(? number?) (string-append (number->string data) " ")]))


(define (flat-def def-lst)
  (if  (empty? def-lst)
      (list)
      (begin
        (let ([f (car def-lst)])
          (if (not (list? f))
              (cons (toString-with-space f) (flat-def (drop def-lst 1)))
              (append (list "(") (flat-def f) (flat-def (drop def-lst 1)) (list ")"))))))) 

(define (lstStr->lstChars lst-str)
  (for/fold ([l empty])
     ([el (in-list lst-str)])
     (append  l (string->list el))))


(define flat (flat-def ' (define b1 :: bool (=> (and (= X x) (= Y y)) (and (= Y y) (= X x))))))
(set! flat (append  (list "\"" "(") flat (list  ")" "\""))) 
(set! flat (lstStr->lstChars flat))
(set! flat (list->string flat))
(display flat)