Comments (3)
Also, I'm a few days away from making a fix. If you are happy I'll make a pull request later this week.
from decorator.
I am sorry, but this is still a won't fix. You have discovered an error of the wrapper technology, which is mixing up default arguments and keyword arguments. The decorator module is doing the right thing, by keeping them separated. The proof:
import inspect
def foo(a, b, c=-1, d=-2, **kwargs): pass
print inspect.getargspec(foo)
ArgSpec(args=['a', 'b', 'c', 'd'], varargs=None, keywords='kwargs', defaults=(-1, -2))
c
and d
are not keyword arguments and must stay in the args
group. The decorator module is doing the right thing, the wrapper approach is failing. The decorator module was designed to fix this error.
from decorator.
I was baffled by this problem as well. Here's what I wanted to do:
def decorate_a_func(path):
def _decorate(func, *args, **kwargs):
baz = kwargs.get('baz') # This didn't work as expected
if baz:
pass
return func(*args, *kwargs)
return decorator(_decorate)
@decorate_a_func(path=PATH)
def do_a_thing(foo, bar, baz):
pass
do_a_thing(
foo=123,
bar=456,
baz=True,
)
At the highlighted line above, kwargs was empty, and args was equal to (123, 456, True)
One solution is to do the following:
baz = args[2]
This is probably fine if you don't mind the decorator being tightly coupled with the function(s) that it decorates, since now the decorator depends on that argument always being in the third position. That doesn't seem very ideal to me.
Alternatively, you can use inspect to build a dictionary of named arguments:
import inspect
def decorate_a_func(path):
def _decorate(func, *args, **kwargs):
func_spec = inspect.getargspec(func)
named_args = dict(zip(func_spec.args, args))
baz = named_args.get('baz')
if baz:
pass
return func(*args, *kwargs)
return decorator(_decorate)
That works as expected, but seems like overkill.
Is there a better solution to this scenario?
from decorator.
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from decorator.