Python - 将函数传递给其他函数

Question

我在将 crapsRoll() 传递到 gameCraps() 时遇到问题。当我定义 gameCraps() 时，出现错误:

Redefining name 'crapsRoll' from outer scope.

在打印语句中我收到一个错误，上面写着

crapsRoll has no value.

import random 

#===========================crapsRoll()===============================
def crapsRoll():
  roll = random.randint(2,12)
  return(roll)

#===========================gameCraps()===============================
def gameCraps(crapsRoll):
  crapsRoll = crapsRoll
  if (crapsRoll == 7 or crapsRoll == 11):
    gameState = 1
  elif (crapsRoll == 2 or crapsRoll == 3 or crapsRoll == 12):
    gameState = 2
  else:
    gameState = 3
  return(gameState)

print(gameCraps())

Answer 1

您不需要将其作为参数传递。只需从另一个函数调用它即可:

def gameCraps():
  crapsRoll = crapsRoll()
  if (crapsRoll == 7 or crapsRoll == 11):
    gameState = 1
  elif (crapsRoll == 2 or crapsRoll == 3 or crapsRoll == 12):
    gameState = 2
  else:
    gameState = 3
  return(gameState)