Python - empty argument in function

Question

I have a function in my script that takes two arguments and save result to file:

def listing(val, ftype):

    with open(sys.argv[3], "a+") as myfile:
        myfile.write(val + ftype + '\n')
    return

but I want to use it when only one argument is provided. I can define argument as empty string '' like that:

listing('', '\n[*]Document files found:\n')

Is it the only way to deal with empty arguments?

If the argument is optional, the function declaration should say so: def listing(ftype, val=None). How the function then deals with the value internally is up to you. — deceze
– deceze ♦, Commented Nov 8, 2017 at 15:05

Daniel Isaac · Accepted Answer · 2017-11-08 15:36:15Z

2

You could use default arguments for this purpose

Using Default Arguments

def listing(val, ftype=''):

    with open(sys.argv[3], "a+") as myfile:
       myfile.write(val + ftype + '\n')
    return

Calling : listing('\n[*]Document files found:\n') # need not give second argument

You could also make this a more generic function by taking variable number of arguments

def listing(*args):
     with open(sys.argv[3], "a+") as myfile:
          myfile.write("".join(args) + '\n')
     return

Calling: listing('String1','String2','String3')

edited Nov 8, 2017 at 15:36

answered Nov 8, 2017 at 15:13

Daniel Isaac

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2 Comments

Jean-François Fabre Over a year ago

then myfile.write("".join(args) + '\n') will be better than the horrible loop

imperato Over a year ago

Thanks, generic vesion is the best for my needs, as I realized that I'm going to reuse this function even with more than 2 arguments, again thanks a lot!

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