为什么我们需要装饰器中的`* args`？

Question

def pass_thru(func_to_decorate): 
    def new_func(*args, **kwargs): #1 
     print("Function has been decorated. Congratulations.") 
     # Do whatever else you want here 
     return func_to_decorate(*args, **kwargs) #2 
    return new_func 


def print_args(*args): 
    for arg in args: 
     print(arg) 


a = pass_thru(print_args) 
a(1,2,3) 

>> Function has been decorated. Congratulations. 
1 
2 
3 

我知道*args在＃1中使用，因为它是一个函数声明。但为什么有必要在＃2中编写*args，即使它不是函数声明？

Answer 1

当函数声明使用的，*args转位置参数成元组：

def foo(a, *args): 
    pass 
foo(1, 2, 3) # a = 1, args = (2, 3) 

当在函数调用中使用，*args扩展元组成的位置参数：

def bar(a, b, c): 
    pass 
args = (2, 3) 
foo(1, *args) # a = 1, b = 2, c = 3 

这是两个相反的过程，所以将它们结合起来可以将任意数量的参数传递给装饰函数。

@pass_thru 
def foo(a, b): 
    pass 
@pass_thru 
def bar(a, b, c): 
    pass 

foo(1, 2)  # -> args = (1, 2) -> a = 1, b = 2 
bar(1, 2, 3) # -> args = (1, 2, 3) -> a = 1, b = 2, c = 3