Python是否可以确定访问方法的对象的类 [英] Can Python determine the class of an object accessing a method

问题描述

反正有做这样的事情吗?

Is there anyway to do something like this:

class A:
    def foo(self):
        if isinstance(caller, B):
           print "B can't call methods in A"
        else:
           print "Foobar"
class B:
    def foo(self, ref): ref.foo()

class C:
    def foo(self, ref): ref.foo()


a = A();
B().foo(a)    # Outputs "B can't call methods in A"
C().foo(a)    # Outputs "Foobar"

A中的 caller 使用某种自省形式来确定调用方法的对象的类吗?

Where caller in A uses some form of introspection to determine the class of the calling method's object?

编辑:

最后，我根据一些建议将它们放在一起:

In the end, I put this together based on some of the suggestions:

import inspect
...
def check_caller(self, klass):
    frame = inspect.currentframe()
    current = lambda : frame.f_locals.get('self')
    while not current() is None:
        if isinstance(current(), klass): return True
        frame = frame.f_back
    return False

由于提供的所有原因，它并不完美，但是感谢您的答复:它们是一个很大的帮助.

It's not perfect for all the reasons supplied, but thanks for the responses: they were a big help.

推荐答案

假设调用者是一个方法，那么可以，通过查看上一帧，然后从本地人中选择self，可以.

Assuming the caller is a method, then yes you can, by looking in the previous frame, and picking out self from the locals.

class Reciever:
    def themethod(self):
        frame = sys._getframe(1)
        arguments = frame.f_code.co_argcount
        if arguments == 0:
            print "Not called from a method"
            return
        caller_calls_self = frame.f_code.co_varnames[0]
        thecaller = frame.f_locals[caller_calls_self]
        print "Called from a", thecaller.__class__.__name__, "instance"

Üglŷ挺不错的，但是行得通.现在，为什么您要执行此操作完全是另一个问题，我怀疑还有更好的方法.不允许将A的整个概念称为B的错误.

Üglŷ as heck, but it works. Now why you would want to do this is another question altogether, I suspect that there is a better way. The whole concept of A isn't allowed to call B is likely to be a mistake.

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