Python装饰器是如何获取被装饰函数的参数的？

Question

def debug(func):
    def wrapper(*args, **kwargs):  
        print "[DEBUG]: enter {}()".format(func.__name__)
        print 'Prepare and say...',
        return func(*args, **kwargs)
    return wrapper  

@debug
def say(something):
    print "hello {}!".format(something)

如题，debug怎么获取something呢？debug只接收了func啊？连带参数一起接收了吗？

Answer 1

# debug 返回的是函数wrapper 所以
@debug
def say(something):
    print "hello {}!".format(something)
# 等价于
debug(say)(something)
# 等价于
wrapper(something)
# 从 def wrapper(*args, **kwargs)可知，something传递到了*args中

Answer 2

say = debug(say) ==> say = warpper
say(something) 等同于 warpper(something)