从方法中确定定义 class

Question

以下Python3.5代码：

class Base(object):
    def __init__(self):
        print("My type is", type(self))

class Derived(Base):
    def __init__(self):
        super().__init__()
        print("My type is", type(self))

d = Derived()

打印：

My type is <class '__main__.Derived'>
My type is <class '__main__.Derived'>

我想知道，在每个 __init__() 中，定义方法的 class，而不是派生 class。所以我会得到以下打印：

My type is <class '__main__.Base'>
My type is <class '__main__.Derived'>

Answer 1

解决方案 1

使用super().__thisclass__:

class Base(object):
    def __init__(self):
        print("My type is", super().__thisclass__)

class Derived(Base):
    def __init__(self):
        super().__init__()
        print("My type is", super().__thisclass__)

d = Derived()

My type is <class '__main__.Base'>
My type is <class '__main__.Derived'>

解决方案 2

不那么优雅，但 hard-wiring class 有效：

class Base(object):
    def __init__(self):
        print("My type is", Base)

class Derived(Base):
    def __init__(self):
        super().__init__()
        print("My type is", Derived)

d = Derived()

输出：

My type is <class '__main__.Base'>
My type is <class '__main__.Derived'>

从方法中确定定义 class

Determine the defining class from within a method

python

inheritance

python-3.5

解决方案 1

解决方案 2