python3：如何组合2个函数装饰器？

Question

Python 3.9： 定义了 2 个函数装饰器。 目标是检测函数是否应用了 0、1 或 2 个装饰器。

为什么下面的代码为第二个装饰器返回“False”？

def decorator1(f):
    def wrapped(*args, **kwargs):
        f(*args, **kwargs)

    wrapped.dec1 = True
    return wrapped


def decorator2(f):
    def wrapped(*args, **kwargs):
        f(*args, **kwargs)

    wrapped.dec2 = True
    return wrapped


@decorator1
@decorator2
def myfunc():
    print(f"running myfunc")


if __name__ == "__main__":
    myfunc()

    print(f"myfunc has decorator1: {getattr(myfunc, 'dec1', False)}")
    print(f"myfunc has decorator2: {getattr(myfunc, 'dec2', False)}")

结果：

running myfunc
myfunc has decorator1: True 
myfunc has decorator2: False

Answer 1

此代码：

@decorator1
@decorator2
def myfunc():
    print(f"running myfunc")

相当于：

def myfunc():
    print('running myfunc')

myfunc = decorator2(myfunc)
myfunc = decorator1(myfunc)

请记住，您将这些动态属性添加到

wrapped

函数中。包装的函数是这两个装饰器中的两个不同的对象：

myfunc = decorator2(myfunc)
print(myfunc)   # <function decorator2.<locals>.wrapped at 0x1028456c0>
myfunc = decorator1(myfunc)
print(myfunc)   # <function decorator1.<locals>.wrapped at 0x102845760>

堆叠装饰器后，您可以预期

decorator1.<locals>.wrapped

具有

dec1

，但它没有

dec2

。就在

decorator2.<locals>.wrapped

:

def decorator1(f):
    def wrapped(*args, **kwargs):
        return f(*args, **kwargs)
    wrapped.dec1 = True
    return wrapped


def decorator2(f):
    def wrapped(*args, **kwargs):
        return f(*args, **kwargs)
    wrapped.dec2 = True
    return wrapped


def myfunc():
    print('running myfunc')


myfunc = decorator2(myfunc)
myfunc = decorator1(myfunc)


print(f"myfunc has decorator1: {getattr(myfunc, 'dec1', False)}")
print(f"myfunc has decorator2: {getattr(myfunc.__closure__[0].cell_contents, 'dec2', False)}")