提问人：Jenson 提问时间：8/4/2023 最后编辑：quamranaJenson 更新时间：8/9/2023 访问量：138
在 Python 中创建一个简单的锁和键函数
Create a simple lock and key function in Python
问：
class Lock:
def __init__(self):
self.key_code = "1234" # Change this to your desired key code
self.is_locked = True
def lock(self):
self.is_locked = True
def unlock(self, entered_code):
if entered_code == self.key_code:
self.is_locked = False
else:
print("Incorrect key code. Lock remains locked.")
def is_locked(self):
return self.is_locked
def main():
my_lock = Lock()
print("The lock is currently locked.")
while my_lock.is_locked():
entered code = input("Enter the key code: ")
my_lock.unlock(entered code)
print("The lock is now unlocked.")
if __name__ == "__main__":
main()
我写了这段代码，但它显示了一个
Type Error: 'bool' object is not callable
The lock is currently locked.
我一直在尝试创建一个锁定机制，但它显示类型错误。我尝试了很多方法，但我仍然无法解决它。
答：
1赞
JaySean
8/4/2023
#1
你快到了：你的代码有轻微的错误 i. 构造函数方法应命名为 init（带有双下划线）而不是 init。 ii. 您同时具有一个名为 is_locked 的实例变量和一个名为 is_locked 的方法。这可能会导致冲突。我将方法重命名为 is_locked_status 以避免冲突。 iii. 调用 is_locked_status 方法时，需要使用括号 （） 调用该方法并获取其返回值。
class Lock:
def __init__(self): # Use double underscores for the constructor method name
self.key_code = "1234"
self.is_locked = True
def lock(self):
self.is_locked = True
def unlock(self, entered_code):
if entered_code == self.key_code:
self.is_locked = False
else:
print("Incorrect key code. Lock remains locked.")
def is_locked_status(self): # Rename the method to avoid name conflict
return self.is_locked
def main():
my_lock = Lock()
print("The lock is currently locked.")
while my_lock.is_locked_status(): # Call the renamed method
entered_code = input("Enter the key code: ")
my_lock.unlock(entered_code)
print("The lock is now unlocked.")
if __name__ == "__main__":
main()
评论
0赞
chepner
8/4/2023
is_locked最好用作谓词方法的名称。由于没有人应该直接访问 instance 属性（即，在不使用方法的情况下进行设置），因此 instance 属性最好命名（或者实际上只是）。my_lock.is_locked = Falseunlock_is_locked_locked
1赞
rkochar
8/4/2023
#2
这里有 2 种方法可以做到这一点。
class Lock:
def __init__(self):
self.key_code = "1234" # Change this to your desired key code
self.is_locked = True
def lock(self):
self.is_locked = True
def unlock(self, entered_code):
if entered_code == self.key_code:
self.is_locked = False
else:
print("Incorrect key code. Lock remains locked.")
def is_locked(self):
return self.is_locked
def main():
my_lock = Lock()
print("The lock is currently locked.")
while my_lock.is_locked: # <-- Use variable here instead of getter
entered_code = input("Enter the key code: ")
my_lock.unlock(entered_code)
print("The lock is now unlocked.")
if __name__ == "__main__":
main()
这可行，但您想使用 getter 来访问您的变量。
class Lock:
def __init__(self):
self.key_code = "1234" # Change this to your desired key code
self.locked = True
def lock(self):
self.locked = True
def unlock(self, entered_code):
if entered_code == self.key_code:
self.locked = False
else:
print("Incorrect key code. Lock remains locked.")
def is_locked(self):
return self.locked
def main():
my_lock = Lock()
print("The lock is currently locked.")
while my_lock.is_locked():
entered_code = input("Enter the key code: ")
my_lock.unlock(entered_code)
print("The lock is now unlocked.")
if __name__ == "__main__":
main()
问题在于变量和方法具有相同的名称。提示在错误中 - 它使用变量认为它是一个方法，而不是使用具有相同名称的方法。更改变量以使其正常工作。locked
评论