遍历堆栈应该会给出答案。
理想情况下,答案应该是在调用者的堆栈帧中。
问题是,堆栈帧只记录函数
名称(例如:“f”、“g”、“h”等)有关的任何信息
课程丢失。试图对丢失的信息进行逆向工程,
通过导航类层次结构(与
堆栈帧),并没有让我走得太远,并且变得复杂。
因此,这是一种不同的方法:
将类信息注入栈帧
(例如使用局部变量),
并从被调用的函数中读取它。
import inspect
class A:
def f(self):
frame = inspect.currentframe()
callerFrame = frame.f_back
callerLocals = callerFrame.f_locals
return callerLocals['cls']
class B(A):
def g(self):
cls = B
return self.f()
def f(self):
cls = B
return super().f()
class C(B):
def h(self):
cls = C
return super(B, self).f()
def f(self):
cls = C
return super().f()
c = C()
assert c.h() == C
assert c.g() == B
assert c.f() == B
Related:
从检查堆栈获取完全限定方法名称 https://stackoverflow.com/questions/16589547/get-fully-qualified-method-name-from-inspect-stack
不修改子类的定义:
添加了一个“外部”装饰器来包装类方法。
(至少作为临时解决方案。)
import inspect
class Injector:
def __init__(self, nameStr, valueStr):
self.nameStr = nameStr
self.valueStr = valueStr
# Should inject directly in f's local scope / stack frame.
# As is, it just adds another stack frame on top of f.
def injectInLocals(self, f):
def decorate(*args, **kwargs):
exec(f'{self.nameStr} = {self.valueStr}')
return f(*args, **kwargs)
return decorate
class A:
def f(self):
frame = inspect.currentframe()
callerDecoratorFrame = frame.f_back.f_back # Note:twice
callerDecoratorLocals = callerDecoratorFrame.f_locals
return callerDecoratorLocals['cls']
class B(A):
def g(self): return self.f()
def f(self): return super().f()
class C(B):
def h(self): return super(B, self).f()
def f(self): return super().f()
bInjector = Injector('cls', B.__name__)
B.g = bInjector.injectInLocals(B.g)
B.f = bInjector.injectInLocals(B.f)
cInjector = Injector('cls', C.__name__)
C.h = cInjector.injectInLocals(C.h)
C.f = cInjector.injectInLocals(C.f)
c = C()
assert c.h() == C
assert c.g() == B
assert c.f() == B
我发现这个链接很有趣
(但没有利用这里的元类):
python 中的元类是什么 https://stackoverflow.com/questions/100003/what-are-metaclasses-in-python/6581949#6581949
也许有人甚至可以替换函数定义*,
其代码与原始代码重复的函数;
但直接在其范围内添加了当地人/信息。
*
也许在类定义完成之后;
也许在类创建期间(使用元类)。