Preserve Signature in Decorator python 2

问题

I am writing a decorator which will catch TypeError for incorrect number of arguments in a function call and will print a customised message. The code is here:

import inspect

def inspect_signature(f):
    def decorate(*args, **kwargs):
        try:
            f(*args, **kwargs)
        except TypeError as e:
            print('Failed to call "{}" with signature {}. Provided args={} and kwargs={}.'.format(
                f.__name__, inspect.getargspec(f).args, args, kwargs))
        return f
    return decorate


@inspect_signature
def func(foo, bar):
    pass
func('a')
func('a', 'b')

I get the following output:

Failed to call "func" with signature ['foo', 'bar']. Provided args=('a',) and kwargs={}.
Called successfully with foo=a, bar=b
ArgSpec(args=[], varargs='args', keywords='kwargs', defaults=None)

The function signature is empty. Please suggest me a solution how can I retain it?

PS: I am using python2 and cannot switch to python3.

回答1:

You missed this here. func(*foo, **bar)

In your case func('a') was not working as you gave fixed arg for it.

You need to pass a variable number of arguments to your function

@inspect_signature
def func(*foo, **bar):
    pass

回答2:

You can refer to Preserving signatures of decorated functions.

In a word, you can use decorator module in Python2 and Python3. When you use Python3.4+, you can use inspect.wraps to Preserving signatures of decorated functions.

If you don't want to use decorator module, you can use eval to make a decorator. In General, eval is unpopular, so, decorator module may be the best solution in Python2.

来源：https://stackoverflow.com/questions/42359024/preserve-signature-in-decorator-python-2