每次函数运行时增加计数

Question

我刚开始学Python。我正在翻阅一本书，做一个基于文本的游戏。你知道吗

所以我有一个房间。我想让玩家死，如果他/她进入房间3次，但不知道怎么做。你知道吗

def spawn():
    count = 0
    count += 1
    print(count)
    print("You dropped down nearly to the magma.")
    print("There are four doors around you.")
    print("Which one do you take?")
    ch = input("Top, bottom, left or right? > ")
    if count = 4:
        dead("You wandered around too much and died.")
    else:
        print()

我试着用印刷来追踪数字，但我不能使它增加。我做错什么了？你知道吗

编辑：当我把count放在函数外时，它给出：

Traceback (most recent call last):
  File "ex.py", line 147, in <module>
    spawn()
  File "ex.py", line 14, in spawn
    count += 1
UnboundLocalError: local variable 'count' referenced before assignment

Answer 1

或者你可以这样做：

def spawn():
    if not hasattr(spawn, 'count'):
        spawn.count = 0
    spawn.count += 1
    print(spawn.count)
    print("You dropped down nearly to the magma.")
    print("There are four doors around you.")
    print("Which one do you take?")
    ch = input("Top, bottom, left or right? > ")
    if spawn.count == 4:
        dead("You wandered around too much and died.")
    else:
        print()

Answer 2

在函数中，每次都将一个局部变量设置为0，这意味着函数完成后，这个变量就不存在了。你知道吗

诀窍是在函数存在后使用一个保持“活动”的变量。例如，函数外的变量，或者可以向递增的函数添加属性。后者的优点是，更清楚的是，这与功能有关，例如：

def spawn():
    spawn.count += 1
    print(spawn.count)
    print("You dropped down nearly to the magma.")
    print("There are four doors around you.")
    print("Which one do you take?")
    ch = input("Top, bottom, left or right? > ")
    if spawn.count == 4:
        dead("You wandered around too much and died.")
    else:
        print()

spawn.count = 0

注意，您还忘记了对if语句使用双等号（==）（相等检查与赋值）。你知道吗