需要帮助调用我的刽子手游戏中的函数吗

Question

需要在我的python游戏中调用函数的帮助

我正在使用turtle开发我的第一个Python刽子手游戏。我只需要找出一种方法，在用户输入错误字符时调用下面的函数

def drawHangman(counter):

  def drawNoose():
    turtle.speed(10)
    turtle.color("Black")
    turtle.forward(120)
    turtle.forward(-60)
    turtle.left(90)
    turtle.forward(150)
    turtle.right(90)
    turtle.forward(100)
    turtle.right(90)
    turtle.forward(30)
    turtle.right(90)

  def drawHead():
    turtle.circle(15)
    turtle.circle(15, 180) # draw a semicircle
    turtle.right(90)

  def drawArms():
    turtle.forward(5)
    turtle.left(90)
    turtle.forward(20)
    turtle.forward(-40)
    turtle.forward(20)
    turtle.right(90)

  def drawTorso():
    turtle.forward(30)

  def drawLegs():
    turtle.left(45)
    turtle.forward(30)
    turtle.forward(-30)
    turtle.right(90)
    turtle.forward(30)
    turtle.forward(-30)
    turtle.left(45)

  if counter==0:
    drawNoose()
  elif counter==1:
    drawNoose()
    drawHead()
  elif counte)
  elif counter==4:
    drawLegs()r==2:
    drawArms()
  elif counter==3:
    drawTorso)
  elif counter==4:
    drawLegs()()
  elif counter==4:
    drawLegs()

有人知道我该怎么做吗

我的代码如下：

guesses = ''

turns = 6

while turns > 0:


    failed = 0


    for char in word:


        if char in guesses:
            print(char)

        else:
            print("_")



            failed +=1



    if failed == 0:

        print("You Win")


        print("The word is: ", word)
        break


    guess = input("guess a character:")


    guesses += guess


    if guess not in word:

        turns -= 1


        print("Wrong")





        print("You have", + turns, 'more guesses')

        if turns == 0:
            print("You Loose")

Answer 1

你使用csn吗

def drawHangman(counter):

  def drawNoose():
    turtle.speed(10)
    turtle.color("Black")
    turtle.forward(120)
    turtle.forward(-60)
    turtle.left(90)
    turtle.forward(150)
    turtle.right(90)
    turtle.forward(100)
    turtle.right(90)
    turtle.forward(30)
    turtle.right(90)

  def drawHead():
    turtle.circle(15)
    turtle.circle(15, 180) # draw a semicircle
    turtle.right(90)

  def drawArms():
    turtle.forward(5)
    turtle.left(90)
    turtle.forward(20)
    turtle.forward(-40)
    turtle.forward(20)
    turtle.right(90)

  def drawTorso():
    turtle.forward(30)

  def drawLegs():
    turtle.left(45)
    turtle.forward(30)
    turtle.forward(-30)
    turtle.right(90)
    turtle.forward(30)
    turtle.forward(-30)
    turtle.left(45)

  if counter==0:
    drawNoose()
  elif counter==1:
    drawNoose()
    drawHead()
  elif counter==2:
    drawArms()
  elif counter==3:
    drawTorso()
  elif counter==4:
    drawLegs()

此函数（与示例链接相同）

然后像drawHangman()一样调用这个函数，您可以在代码的第86行使用它

还有一个failed变量；您可以将此变量放入函数drawHangman()的输入中，就像drawHangman(failed)