CODEWITHSUNDEEP
pythondecoratorpython-decorators
lajarre
lajarre

Reputation: 5162

Python decorator module - kwargs

Using the decorator module from Michele Simionato, I'm getting a weird behavior with respect to kwargs. Here's an example using the simple example that is in the documentation:

def _trace(f, *args, **kw):
    kwstr = ', '.join('%r: %r' % (k, kw[k]) for k in sorted(kw))
    print("calling %s with args %s, {%s}" % (f.__name__, args, kwstr))
    return f(*args, **kw)

def trace(f):
    return decorate(f, _trace)

@trace
def f(a, b, c=1, d=1):
    return a + b + c + d

f(1,2,c=2,d=2)

prints calling f with args (1, 2, 3, 4), {}

instead of calling f with args (1, 2), {'c': 2, 'd': 2}

Is it wanted behavior? Why so?

Or did I get something wrong?

Upvotes: 0

Views: 112

Answers (2)

iBelieve
iBelieve

Reputation: 1544

This seems to be the expected behavior. From the page you linked:

>>> @trace
... def f(x, y=1, z=2, *args, **kw):
...     pass

>>> f(0, 3)
calling f with args (0, 3, 2), {}

>>> print(getargspec(f))
ArgSpec(args=['x', 'y', 'z'], varargs='args', varkw='kw', defaults=(1, 2))

Notice how the y and z keyword arguments show up as regular args, not keyword args.

Upvotes: 1

AlokThakur
AlokThakur

Reputation: 3741

Expectation is not correct with current prototype of f(a, b, c=1, d=1), It is just default argument type. Change f to accept keyargs- 

@trace
def f(a, b, **kargs):
    return a + b + kargs['c'] + kargs['d'] 
f(1,2,c=2,d=2)
calling f with args (1, 2), {'c': 2, 'd': 2}

Upvotes: 0

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