函數不調用另一個函數

Question

我有一個函數作爲教程遊戲的一部分。如果條件滿足（如果對象==「代碼」），則有問題的功能應觸發另一功能

# right room 
def right_room(): 
    print "You see a table with two objects: a map and a code translator" 
    print "You can take one object" 
    print "Which object do you take?" 

    next = raw_input("> ") 

    if "map" in next and "code" in next: 
     dead("You're greed surpassed your wisdom.") 
    elif "map" in next: 
     print "OK, you have the map." 
     theobject = "map" 
     print "Now you must exit and go ahead" 
     return theobject 
     opening() 

    elif "code" in next: 
     print "OK, you have the code." 
     theobject = "code" 
     print "Now you must exit and go ahead." 
     return theobject 
     opening() 

但是打開不被稱爲？這裏是輸出：

你在Labrynthe。你左邊有一扇門。你的權利 有一個門。或者你可以繼續。

right你看到一個包含兩個對象的表：一個地圖和一個代碼翻譯器你可以帶一個對象你拿哪個對象？ 代碼確定，你有代碼。現在你必須退出並繼續。然後

上面的函數是指人在送他們回到起點和提示輸入「超前」，在終端：

# opening scene 
def opening(): 
    print "You're in a Labrynthe." 
    print "There's a door on your left." 
    print "There's a door on your right." 
    print "Or you can go ahead." 

    next = raw_input("> ") 

    if "right" in next: 
     right_room() 
    elif "left" in next: 
     left_room() 
    elif "ahead" in next: 
     ahead() 
    else: 
     print "Which way will you go?" 

但開口（）沒有被調用。相反，Python似乎完成腳本並退出。

Answer 1

Python中的return語句（以及幾乎所有的語言 - Haskell的顯着例外 - ）意味着函數應在處停止。如果語句是return expr那麼表達式的值可供調用者使用。否則，在Python中，調用者可以使用值None。

因此，您可能需要在return語句之前移動函數調用。