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This topic created in 2596 days ago, the information mentioned may be changed or developed.
现实里真的会遇到这种情况...之前从没考虑过,.
Supplement 1 · Mar 28, 2019
如果我装饰器没写错,答案是会报错. 目前正在尝试把wrapper写成类.
def desc(cls):
def wrapper(*args, **kwargs):
print(123)
return cls(*args, **kwargs)
return wrapper
@desc
class FOOA(object):
pass
class FOOB(FOOA):
pass
f = FOOB()
Supplement 2 · Mar 28, 2019
标准答案是,取决于你的装饰器是怎么写的.
```
def desc(cls):
class wrapper(object):
def __new__(cls2, *args, **kwargs):
print (123)
cls.a = 111
return object.__new__(cls, *args, **kwargs)
return wrapper
@desc
class FOOA(object):
pass
class FOOB(FOOA):
pass
f = FOOB()
print (f.a)
```
```
def desc(cls):
class wrapper(object):
def __new__(cls2, *args, **kwargs):
print (123)
cls.a = 111
return object.__new__(cls, *args, **kwargs)
return wrapper
@desc
class FOOA(object):
pass
class FOOB(FOOA):
pass
f = FOOB()
print (f.a)
```
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1
chaleaochexist OP 晚上下班回家我会试试,
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2
Trim21 Mar 28, 2019
应该是不会, 类装饰器只是把这个类在定义的时候处理了一遍, 子类继承到的是被处理之后的类, 这里可能会受一定的影响, 而不会继承装饰器.
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3
yushenglin Mar 28, 2019
第一个装饰器相当于返回了一个函数，类不能继承函数类型，肯定会报错呀，第二个相当于重写了类的__new__（）函数，下面的类继承了，肯定会有影响呀
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