Python装饰器: TypeError: function takes 1 positional argument but 2 were given

Question

我正在尝试测试 Python 中装饰器的实用性。当我编写以下代码时，出现错误:

TypeError: fizz_buzz_or_number() takes 1 positional argument but 2 were given

我首先定义一个函数log_calls(fn)为

def log_calls(fn):
    ''' Wraps fn in a function named "inner" that writes
    the arguments and return value to logfile.log '''
    def inner(*args, **kwargs):
        # Call the function with the received arguments and
        # keyword arguments, storing the return value
        out = fn(args, kwargs)

        # Write a line with the function name, its
        # arguments, and its return value to the log file
        with open('logfile.log', 'a') as logfile:
            logfile.write(
               '%s called with args %s and kwargs %s, returning %s\n' %
                (fn.__name__,  args, kwargs, out))

        # Return the return value
        return out
    return inner

之后，我使用 log_calls 将另一个函数装饰为:

@log_calls
def fizz_buzz_or_number(i):
    ''' Return "fizz" if i is divisible by 3, "buzz" if by
        5, and "fizzbuzz" if both; otherwise, return i. '''
    if i % 15 == 0:
        return 'fizzbuzz'
    elif i % 3 == 0:
        return 'fizz'
    elif i % 5 == 0:
        return 'buzz'
    else:
        return i

当我运行以下代码时

for i in range(1, 31):
    print(fizz_buzz_or_number(i))

出现错误 TypeError: fizz_buzz_or_number() Takes 1 Positional argument but 2 were Gived 。

我不知道这个装饰器有什么问题，以及如何解决这个问题。

Answer 1

您将在此处向您的修饰函数传递 2 个参数:

out = fn(args, kwargs)

如果您想申请args元组和 kwargs字典作为变量参数，回显函数签名语法，因此使用 *和**再次:

out = fn(*args, **kwargs)

请参阅Call expressions reference documentation :

If the syntax *expression appears in the function call, expression must evaluate to an iterable. Elements from these iterables are treated as if they were additional positional arguments.

[...]

If the syntax **expression appears in the function call, expression must evaluate to a mapping, the contents of which are treated as additional keyword arguments.