How can I use default arguments in a method with python-fire?

Question

I am having an issue with python-fire when a method has arguments with default values. Consider the following code:

import fire

class SomeClass(object):
    def __init__(self):
        self.a = ''
        self.b = ''

    def methoda(self):
        self.a = 'A'
        return self

    def methodb(self, x='B123'):
        self.b = self.a + x
        return self

    def __str__(self):
        return self.b

if __name__ == '__main__':
    s = SomeClass()
    s.methodb().methoda()
    print(s.b) # this prints B123 correctly

    fire.Fire(SomeClass)

As seen in the comments, in the print(s.b), it is printingB123` correctly. But when i change the order of the methods being called in fire from the command line, I am getting the odd behavior.

Example:

> python x.py
B123 # correct

> python x.py methoda methodb
Here: B123
AB123 # both correct

> python x.py methodb --x B123 methoda
Here: B123
B123 # again both correct

> python x.py methodb methoda
Here: B123
methoda # this is not correct. It should print B123 here also

As you can see with the last example, if i call methodb (which has an argument with a default value), it prints methoda instead of B123 as expected.

My question is, How can I use a method that has a default argument value first in this type of scenario without passing in --x=something?

In short, how can i make > python x.py methodb methoda properly print B123?

Answer 1

You need to change def methodb(self, x='B123'): to def methodb(self, *, x='B123'):, which makes x a keyword only argument, so it can no longer be used as a positional argument. This way fire will use the default value.

There has also been a github issue discussing this.