未綁定錯誤，分配

Question

我正在做一個小的文本遊戲的樂趣之前引用局部變量。我想使用位於我稱之爲functionala的函數文件中的函數。

問題，attack()功能，不工作，程序與錯誤崩潰：

Traceback (most recent call last): 
    File "C:\Users\seanm\Desktop\Programming\The mists of Alandria\Mists_of_alandria.py", line 22, in <module> 
    functionala2.attack() 
    File "C:\Users\seanm\Desktop\Programming\The mists of Alandria\functionala2.py", line 27, in attack 
    variablestamina += 2 
UnboundLocalError: local variable 'variablestamina' referenced before assignment 

的functionala文件的新的和改進的版本是什麼似乎會引起問題：

variablestamina = 20 
variablehealth = 40 
variablemonsterhealth = 30 
variableattacktype1 = ("Lightattack") 
variableattacktype2 = ("Mediumattack") 
variableattacktype3 = ("Heavyattack") 

def attack(): 
    variableattackquery = input("You can execute three types of attacks. Lightattack does 2 damage and takes no stamina. Mediumattack does 4 damage and takes 2 stamina. Heavyattack does 7 damage and takes 5 stamina. You can only do one per turn: ") 
    if variableattackquery == variableattacktype1: 
     variablemonsterhealth -= 2 
     variablestamina -= 2 
    if variableattackquery == variableattacktype2: 
     variablemonsterhealth -= 4 
     variablestamina -= 4 
    if variableattackquery == variableattacktype3: 
     variablemonsterhealth -= 7 
     variablestamina -= 7 
    variablestamina += 2 
    variablestamina = min(20, variablestamina) 
    print ("The "+monster+" has "+str(variablemonsterhealth)+" health left") 
    print ("You have "+str(variablestamina)+" stamina left") 
    monsterattack = random.randrange(4,6) 
    variablehealth -= monsterattack 
    print ("The "+monster+" attacks you for "+str(monsterattack)) 
    print ("You have "+str(variablehealth)+" health left") 
    print() 

Answer 1

這似乎這樣做，在一個單一的文件的更清潔的方式。你可能想看看使用類。

從控制檯，調用game()開始遊戲，僅此而已。本場比賽將結束時，無論是怪物，或者您有健康< = 0

代碼：

from random import randrange 

def game(): 
    stamina = 20 
    health = 40 
    monsterhealth = 30 
    monster = 'orc' 
    attacks = {'light':(-2,0),'medium':(-4,-2),'heavy':(-7,-4)} 
    while True: 
     a = input('you can execute 3 types of attacks, light, medium or heavy... pick one.') 
     a = a.lower().strip() 
     if a in attacks: 
      stamina, health, monsterhealth = attack(stamina, health, monsterhealth, monster, attacks[a]) 
      if stamina <= 0: 
       print 'you have died...' 
       break 
      elif monsterhealth <= 0: 
       print 'the {} has died...'.format(monster) 
       break 
     else: 
      break 

def attack(stamina, health, monsterhealth, monster, att): 
    monsterhealth += att[0] 
    stamina += att[1] 
    stamina = min(20, stamina) 
    print('the {} has {} health remaining'.format(monster,monsterhealth)) 
    print('you have {} stamina remaining'.format(stamina)) 
    ma = randrange(4,6) 
    health -= ma 
    print('the {} attacks you for {}'.format(monster,ma)) 
    print('you have {} health left'.format(health)) 
    return stamina, health, monsterhealth 

注意：即使這樣做在一個單一的文件，你需要範圍的變量「主」程序（game），並將它們傳遞給attack函數。否則，提到這些名字將引發同樣的錯誤，你可以重現這個像這樣：

m = 1 
def foo(): 
    m += 1 '## m doesn't exist within scope of foo, so it will raise the same error 

然而，這可能會產生混淆，下面將不引發錯誤：

m = 1 
def foo(): 
    print m 

也不會對這樣的：

m = 1 
def foo(): 
    a = m 
    print a 

但是這兩個似乎有種下鍋y和它通常是更好地從主程序到c值傳遞調用函數/方法/等，並返回適當的值給調用者。