Returning actual argument expression to a function call?

[Steven D'Aprano]

On Fri, 09 Nov 2007 22:03:00 -0800, Paddy wrote:

> Hi,
> # If I have a function definition
...
> # The problem is that for my application to work, 
> # Python newbies would have to write lambda when they 
> # know they are after the result. 

I don't understand why you think this is the case.


> Its my program 
> # that would require the lambda (or def), which 
> # is a distraction from their problem.
...
> Any ideas on implementing f1 so I can do f2?

I've read your post three times now, and I'm still not absolutely sure 
what you're trying to accomplish.

I think you want something like this:

You have a function f1(*args) which returns a result. You want a global 
flag that tells f1 to stick it's arguments in a global variable. Am I 
close?

If so, something like this should work:

# Define someplace to store the arguments.
last_args = (None, None)
# And a flag.
capturecall = True

# A decorator to wrap a function.
def capture_args(func):
    def f(*args, **kwargs):
        if capturecall:
            global last_args
            last_args = (args, kwargs)
        return func(*args, **kwargs)
    f.__name__ = "capturing_" + func.__name__
    return f



Now you can do this:

>>> def f1(x, y=3):
...     return x + y
...
>>> f1 = capture_args(f1)
>>> f1(5)
8
>>> last_args
((5,), {})
>>> @capture_args
... def f2(x):
...     return 2**x
...
>>> f2(0.4+1)
2.6390158215457884
>>> last_args
((1.3999999999999999,), {})



In the second example, if you are trying to capture the expression "0.4
+1", I don't think that is possible. As far as I know, there is no way 
for the called function to find out how its arguments were created. I 
think if you need that, you need to create your own parser.



-- 
Steven.