decorator - python中的带参数装饰器可以调用被装饰的函数中的参数不(除了全局变量)

Question

例子:

def lo(func0, *args0, **kwargs0):
    def _wrapper(func):
        def wrapper(*args, **kwargs):
            func(*args, **kwargs)
            func0(*args0, **kwargs0)
        return wrapper
    return _wrapper


def f1(a):
    print a+1

@lo(f1, a)
def f(a):
    print a

f(1)

如何才能让装饰器lo调用函数f的参数a？

Answer 1

What are you doing? The a in the decorator call is not defined. I don't quite understand what you want to do, see question X-Y.

Do you want this?

def lo(func0):
    def _wrapper(func):
        def wrapper(*args, **kwargs):
            func(*args, **kwargs)
            func0(*args, **kwargs)
        return wrapper
    return _wrapper

def f1(a):
    print(a+1)

@lo(f1)
def f(a):
    print(a)

f(1)