Question

Javascript has a poorly constructed but convenient “arguments” variable inside every function, such that you can pass arguments through a function like so:

function foo(a, b, c) {
    return bar.apply(this, arguments);
}
function bar(a, b, c) {
    return [a, b, c];
}
foo(2, 3, 5);    // returns [2, 3, 5]

Is there an easy way to do a similar thing in Python?

Answer 1

Yeah, this is what I should have said.

def foo(*args):
    return bar(*args)

You don’t need to declare the function with (a,b,c). bar(…) will get whatever foo(…) gets.

My other crummier answer is below:

---

I was so close to answering “No, it can’t easily be done” but with a few extra lines, I think it can.
@cbrauchli great idea using locals(), but since locals() also returns local variables, if we do

def foo(a,b,c):
    n = "foobar" # any code that declares local variables will affect locals()
    return bar(**locals())

we’ll be passing an unwanted 4th argument, n, to bar(a,b,c) and we’ll get an error. To solve this, you’d want to do something like arguments = locals() in the very first line i.e.

def foo(a, b, c):
    myargs = locals() # at this point, locals only has a,b,c
    total = a + b + c # we can do what we like until the end
    return bar(**myargs) # turn the dictionary of a,b,c into a keyword list using **