Question

我刚刚开始学习Python。我正在看一本书，正在做一个基于文本的游戏。

所以我有一个房间。如果他/她进入房间3次，但又不知道该怎么做，我想让他/她死。

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()

我尝试通过打印跟踪数量，但无法使其增加。我在做什么错了？

编辑：当我将count放在函数之外时，它会给出：

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

在函数中，每次将 local 变量设置为0，这意味着之后函数完成，该变量不再存在

诀窍是使用一个在函数存在后仍保持“活动”状态的变量。例如，函数外部的变量，或者您可以在要递增的函数中添加属性。后者的优点是更清楚地表明这与功能有关，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

请注意，您还忘记为==语句（相等性检查与赋值）使用双等号（if）。

Answer 2

或者您可以这样做：

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()

每次函数运行时计数都会增加

2 个答案: