I want to pass a function, f(a=1,b=2) into g and use the 'a' value in g
def f(a,b): pass
def g(f): #print f.a
g(f(1,2)) should result in an output of 1
I looked into the inspect module but can't seem to get hold of f in g
This is as far as my programming knowledge has got me :
def g(f):
print(list(inspect.signature(f).parameters.keys()))
g(f(1,2)) results in: TypeError: None is not a callable object
This is not possible. When you call g(f(1,2)), f(1,2) is finished before g runs. The only way this would be possible would be for f to return its arguments.
You need to call g appropriately, so the f, as a function, is an argument:
g(f, 1, 2)
Now work that into your function:
def g(f, *args):
print(list(inspect.signature(f).parameters.keys()))
... and from here, you can iterate through args to make the proper call to f.
Does that get you moving?
You could do something like this:
def f(a, b):
print(a)
print(b)
Then define wrapper as:
def wrapper(func, a, b):
func(a, b)
Or if you want more flexibility use *args:
def wrapper(func, *args):
func(*args)
That flexibility comes with some risk if the number of arguments don't match. Which means you'll need to take care that all func passed to wrapper have a consistent signature.
You could use **kwargs which would help with the above, then:
def wrapper(func, **kwargs):
func(**kwargs)
Then calls to wrapper would look like:
wrapper(f, a=1, b=2)
This would allow for more flexibility and signature of func could vary as needed.
You could turn wrapper into a decorator but that's another question.
