循環內的函數行爲有所不同

Question

所以我有一個循環來重複我對dota所做的小文本遊戲，但是當函數'play'在循環中被調用時，它不會返回我的cond函數的結果，它只需要一個輸入，然後移動到下一個循環。

;;;;learn the invoker combo's 
(defparameter *invoker-combo* '((cold-snap (3 0 0) 'QQQ);all of the possible invoker combo's 
           (ghost-walk (2 1 0) 'QQW) 
           (Ice-Wall (2 0 1) 'QQE) 
           (EMP (0 3 0) 'WWW) 
           (Tornado (1 2 0) 'QWW) 
           (Alacrity (0 2 1) 'WWE) 
           (Sun-Strike (0 0 3) 'EEE) 
           (Forge-Spirit (1 0 2) 'QEE) 
           (Chaos-Meteor (0 1 2) 'WEE) 
           (Deafening-Blast (1 1 1) 'QWE))) 
(defun rand-combo (invoker-combo);returns a random combo 
    (nth (random (length invoker-combo))invoker-combo)) 

(defun count-letters (input);converts the keyboard strokes into numbers to be compared as it doesn't matter what order they are in, just that there is the correct quantity of them e.g QQE could also be written QEQ. 
    (append 
     (list (count #\Q input) 
       (count #\W input) 
       (count #\E input)))) 

(defun try-for-combo (rand-combo);takes i-p and compares it with the value for the random combo 
    (print(car rand-combo)) 
    (let* ((i-p (string-upcase(read-line))) 
      (try (count-letters i-p))) 
      (cond ((equal try (cadr rand-combo))'Good-job) 
        ((equal i-p "END")(list 'Thanks 'for 'playing)) 
        (t (list i-p 'was 'wrong 'correct 'is (caddr(assoc (car rand-combo)*invoker-combo*))))))) 

(defun play() 
    (try-for-combo (rand-combo *invoker-combo*))) 

(defun loop-play (x) 
    (loop for i from 0 to x 
     :do (play))) 

如果我調用該函數 '玩'，我得到以下O/P：

FORGE-SPIRIT asdf 
("ASDF" WAS WRONG CORRECT IS 'QEE) 

或

ALACRITY wwe 
GOOD-JOB 

但如果我調用該函數 '循環播放' 我得到以下o/p：

Break 3 [7]> (loop-play 2)  
SUN-STRIKE eee  
ALACRITY wwe 
TORNADO qww 
NIL 

有人可以解釋對我來說爲什麼會發生這種情況？ 編輯：隨時更改標題，我真的不知道要放什麼。

Answer 1

您的try-for-combo函數實際上並不是輸出什麼。相反，它返回值。

在REPL中，如果您評估表單，如(+ 1 2)，它將始終在最後打印該表單的評估（在此例中爲3）。但是，請考慮(+ 1 (print 2))。 print函數實際上將參數輸出到標準輸出，然後返回值本身。因此，這將顯示（在REPL）

2 
3 

的2首先輸出，因爲(print 2)本身打印2。然後，表格(+ 1 (print 2))的計算結果與(+ 1 2)或3的計算結果相同。

在你的情況，你的try-for-combo函數應該是這樣的：

(defun try-for-combo (rand-combo) 
    (print (car rand-combo)) 
    (let* ((i-p (string-upcase(read-line))) 
     (try (count-letters i-p))) 
    (print 
    (cond 
     ((equal try (cadr rand-combo)) 'Good-job) 
     ((equal i-p "END") (list 'Thanks 'for 'playing)) 
     (t (list i-p 'was 'wrong 'correct 'is (caddr(assoc (car rand-combo) *invoker-combo*)))))) 
    nil)) 

這將打印cond形式的結果，並返回「零」。

Answer 2

這僅僅是差異的程序將輸出和輸出Lisp的系統確實對每個評價之間：

print打印的東西（一個換行符，然後它的參數）並返回一個值。該值由REPL打印。因此，我們看到輸出兩次：

[3]> (print "3") 

"3" 
"3" 

接下來我們要做的progn好幾撥至print。 REPL打印progn表單的值。前三個串由代碼印刷和最後一個字符串是因爲Lisp的REPL印刷值的印刷：

[4]> (progn (print "1") (print "2") (print "3")) 

"1" 
"2" 
"3" 
"3" 

Answer 3

壓痕和代碼的格式化差。請讓你更容易閱讀代碼。

(defun try-for-combo (rand-combo);takes i-p and compares it with the value for the random combo 
    (print(car rand-combo)) 
    (let* ((i-p (string-upcase(read-line))) 
      (try (count-letters i-p))) 
      (cond ((equal try (cadr rand-combo))'Good-job) ; wrong indent level 
        ((equal i-p "END")(list 'Thanks 'for 'playing)) 
        (t (list i-p 'was 'wrong 'correct 'is (caddr(assoc (car rand-combo)*invoker-combo*))))))) 

• 缺乏
• 錯誤了縮進層次
• 結構代碼不清的s表達式之間的空間
• 不使用內置的文檔功能
• 一些線路太長

更好：

(defun try-for-combo (rand-combo) 
    "takes i-p and compares it with the value for the random combo" ; built in doc 
    (print (car rand-combo)) 
    (let* ((i-p (string-upcase (read-line))) 
     (try (count-letters i-p))) 
    (cond ((equal try (cadr rand-combo))       ; indentation 
      'Good-job) 
      ((equal i-p "END") 
      (list 'Thanks 'for 'playing)) 
      (t 
      (list i-p 'was 'wrong 'correct 'is      ; several lines 
       (caddr (assoc (car rand-combo) 
           *invoker-combo*))))))) 

我會建議使用這其實瞭解一些Lisp語言格式的編輯器。像GNU Emacs的/泥，Clozure CL的鐵杉，LispWorks'編輯器...

如果你不確定格式，你也可以要求Lisp的做到這一點。 CLISP並不好，在格式化，而是要像SBCL或CCL會做：

* (let ((*print-case* :downcase)) 
    (pprint '(defun try-for-combo (rand-combo) 
(print (car rand-combo)) 
        (let* ((i-p (string-upcase (read-line))) 
(try (count-letters i-p))) 
          (cond ((equal try (cadr rand-combo)) 
'Good-job) ((equal i-p "END") 
           (list 'Thanks 'for 'playing)) 
(t (list i-p 'was 'wrong 'correct 'is 
             (caddr (assoc (car rand-combo) 
*invoker-combo*))))))))) 

，你會得到很好的格式化代碼：

(defun try-for-combo (rand-combo) 
    (print (car rand-combo)) 
    (let* ((i-p (string-upcase (read-line))) (try (count-letters i-p))) 
    (cond ((equal try (cadr rand-combo)) 'good-job) 
      ((equal i-p "END") (list 'thanks 'for 'playing)) 
      (t 
      (list i-p 'was 'wrong 'correct 'is 
       (caddr (assoc (car rand-combo) *invoker-combo*))))))) 

由編輯器的Lisp代碼自動縮進您節省了大量工作的。

有用於manual indentation提示。