如何使用`with`语句检查对象是否已创建？

Question

我想确保只在“with”语句中实例化该类。

即。这个没关系：

with X() as x:
 ...

而这不是：

x = X()

我如何确保此类功能？

Answer 1

据我所知，没有直接的方法。但是，在调用对象中的实际方法之前，您可以使用布尔标志来检查是否调用了__enter__。

class MyContextManager(object):

    def __init__(self):
        self.__is_context_manager = False

    def __enter__(self):
        print "Entered"
        self.__is_context_manager = True
        return self

    def __exit__(self, exc_type, exc_value, traceback):
        print "Exited"

    def do_something(self):
        if not self.__is_context_manager:
            raise Exception("MyContextManager should be used only with `with`")

        print "I don't know what I am doing"

与with一起使用时，

with MyContextManager() as y:
    y.do_something()

你会得到

Entered
I don't know what I am doing
Exited

但是，当您手动创建对象并调用do_something时，

x = MyContextManager()
x.do_something()

你会得到

Traceback (most recent call last):
  File "/home/thefourtheye/Desktop/Test.py", line 22, in <module>
    x.do_something()
  File "/home/thefourtheye/Desktop/Test.py", line 16, in do_something
    raise Exception("MyContextManager should be used only with `with`")
Exception: MyContextManager should be used only with `with`

注意：这不是一个可靠的解决方案。在调用任何其他方法之前，有人可以单独直接调用__enter__方法，在这种情况下可能永远不会调用__exit__方法。

如果您不想在每个功能中重复检查，您可以将它设为装饰器，就像这样

class MyContextManager(object):

    def __init__(self):
        self.__is_context_manager = False

    def __enter__(self):
        print "Entered"
        self.__is_context_manager = True
        return self

    def __exit__(self, exc_type, exc_value, traceback):
        print "Exited"

    def ensure_context_manager(func):
        def inner_function(self, *args, **kwargs):
            if not self.__is_context_manager:
                raise Exception("This object should be used only with `with`")

            return func(self, *args, **kwargs)
        return inner_function

    @ensure_context_manager
    def do_something(self):
        print "I don't know what I am doing"

Answer 2

确保在with子句中构造实例没有万无一失的方法，但您可以在__enter__方法中创建一个实例并返回该实例而不是self;这是将分配给x的值。因此，您可以将X视为在__enter__方法中创建实际实例的工厂，例如：

class ActualInstanceClass(object):
    def __init__(self, x):
        self.x = x

    def destroy(self):
        print("destroyed")

class X(object):
    instance = None
    def __enter__(self):

        # additionally one can here ensure that the
        # __enter__ is not re-entered,
        # if self.instance is not None:
        #     raise Exception("Cannot reenter context manager")
        self.instance = ActualInstanceClass(self)

    def __exit__(self, exc_type, exc_value, traceback):
        self.instance.destroy()
        return None

with X() as x:
    # x is now an instance of the ActualInstanceClass

当然这仍然是可重用的，但每个with语句都会创建一个新实例。

当然，您可以手动调用__enter__，或者获取对ActualInstanceClass的引用，但更多的是滥用而不是使用。

---

对于更有气味的方法，X()在调用时确实会创建XFactory实例，而不是X实例;反过来，当用作上下文管理器时，创建ActualX实例，它是X的子类，因此isinstance(x, X)将返回true。

class XFactory(object):
    managed = None
    def __enter__(self):
        if self.managed:
            raise Exception("Factory reuse not allowed")

        self.managed = ActualX()
        return self.managed

    def __exit__(self, *exc_info):
        self.managed.destroy()
        return


class X(object):
    def __new__(cls):
        if cls == X:
            return XFactory()
        return super(X, cls).__new__(cls)

    def do_foo(self):
        print("foo")

    def destroy(self):
        print("destroyed")

class ActualX(X):
    pass

with X() as x:
    print(isinstance(x, X))  # yes it is an X instance
    x.do_foo()               # it can do foo

# x is destroyed

newx = X()
newx.do_foo()  # but this can't,
# AttributeError: 'XFactory' object has no attribute 'do_foo'

您可以更进一步，让XFactory使用X的特殊关键字参数创建一个实际的__new__实例，但我认为它太黑魔法无法使用。< / p>

Answer 3

不幸的是，你不能很干净。

上下文管理器需要使用__enter__和__exit__方法，因此您可以使用此方法在类上分配成员变量以检入代码。

class Door(object):

    def __init__(self, state='closed'):
        self.state = state
        self.called_with_open = False

    # When being called as a non-context manger object,
    # __enter__ and __exit__ are not called.
    def __enter__(self):
        self.called_with_open = True
        self.state = 'opened'

    def __exit__(self, type, value, traceback):
        self.state = 'closed'

    def was_context(self):
        return self.called_with_open


if __name__ == '__main__':

    d = Door()
    if d.was_context():
        print("We were born as a contextlib object.")

    with Door() as d:
        print('Knock knock.')

有状态对象方法有一个很好的附加好处，即能够判断以后调用__exit__方法，或者在以后的调用中干净地处理方法要求：

def walk_through(self):
    if self.state == 'closed':
        self.__enter__
    walk()

Answer 4

到目前为止，所有答案都没有提供（我认为）OP希望直接 （我认为）OP想要这样的事情：

>>> with X() as x:
 ...  # ok

>>> x = X()  # ERROR

Traceback (most recent call last):
  File "run.py", line 18, in <module>
    x = X()
  File "run.py", line 9, in __init__
    raise Exception("Should only be used with `with`")
Exception: Should only be used with `with`

这是我提出的，它可能不是很强大，但我认为它最接近OP的意图。

import inspect
import linecache

class X():

    def __init__(self):
        if not linecache.getline(__file__,
            inspect.getlineno(inspect.currentframe().f_back)
        ).startswith("with "):
            raise Exception("Should only be used with `with`")

    def __enter__(self):
        return self

    def __exit__(self, *exc_info):
        pass

只要with在使用上下文管理器时与X()位于同一行，这将提供与上面显示的完全相同的输出。

Answer 5

这是一个装饰器，可以自动确保方法不在上下文管理器之外调用：

from functools import wraps

BLACKLIST = dir(object) + ['__enter__']

def context_manager_only(cls):
    original_init = cls.__init__
    def init(self, *args, **kwargs):
        original_init(self, *args, **kwargs)
        self._entered = False
    cls.__init__ = init
    original_enter = cls.__enter__
    def enter(self):
        self._entered = True
        return original_enter(self)
    cls.__enter__ = enter

    attrs = {name: getattr(cls, name) for name in dir(cls) if name not in BLACKLIST}
    methods = {name: method for name, method in attrs.items() if callable(method)}

    for name, method in methods.items():
        def make_wrapper(method=method):
            @wraps(method)
            def wrapper_method(self, *args, **kwargs):
                if not self._entered:
                    raise Exception("Didn't get call to __enter__")
                return method(self, *args, **kwargs)
            return wrapper_method
        setattr(cls, name, make_wrapper())

    return cls

这是一个使用它的例子：

@context_manager_only
class Foo(object):
    def func1(self):
        print "func1"

    def func2(self):
        print "func2"

    def __enter__(self):
        print "enter"
        return self

    def __exit__(self, *args):
        print "exit"

try:
    print "trying func1:"
    Foo().func1()
except Exception as e:
    print e

print "trying enter:"
with Foo() as foo:
    print "trying func1:"
    foo.func1()
    print "trying func2:"
    foo.func2()
    print "trying exit:"

这是作为this duplicate question的答案而写的。

Answer 6

OP的问题was believed to be an XY problem和current chosen answer确实（太过分）了。

我真的不知道OP最初的“ X问题”，但我认为动机并不是从字面上阻止“ x = X()分配工作”。相反，这可能会迫使API用户始终使用x作为上下文管理器，以便始终触发其__exit__(...)，这就是将class X设计为首先要成为上下文管理员。至少，这就是把我带到这个问答环节的原因。

class Holder(object):
    def __init__(self, **kwargs):
        self._data = allocate(...)  # Say, it allocates 1 GB of memory, or a long-lived connection, etc.
    def do_something(self):
        do_something_with(self._data)
    def tear_down(self):
        unallocate(self._data)

    def __enter__(self):
        return self
    def __exit__(self, *args):
        self.tear_down()

# This is desirable
with Holder(...) as holder:
    holder.do_something()

# This might not free the resource immediately, if at all
def foo():
    holder = Holder(...)
    holder.do_something()

也就是说，在学习了这里的所有对话之后，我最终还是保持了Holder类的原样，好了，我只是为tear_down()添加了一个文档字符串：

    def tear_down(self):
        """You are expect to call this eventually; or you can simply use this class as a context manager."""
        ...

毕竟，我们都同意这里的成年人...