What's the difference between these two simple codes? More specifically, what is this if statement used for?

Question

I'm reading a book on Python and it's presenting this code:

def myfunc(*args):
    for a in args:
        print(a, end='')
    if args:
        print()

I'm confused what the point of if args is when the code below also shows the same output.

def myfunc(*args):
    for a in args:
        print(a, end='')

Answer 1

Both versions of your function print the list of arguments it receives.

The end='' parameter makes sure that the \n character is not appended for each print, as would happen by default, printing a line for every argument.

In this way you will have an output like this

arg1arg2...argN<noNewline>

(all the arguments are joined with not even a space separating them). With no end='' , instead, you would have had

arg1
arg2
...
argN

---

Since you might want a trailing newline, in the end you call print() , which just prints a newline. But only if the args list is not empty ( if args: ), in order to avoid a "strange" empty line even when no arguments are present.

Answer 2

if args will be True if any arguments exist.

print() just prints a blank line.

Answer 3

*args allows you forward as many arguments to a function as possible. *args always returns set . You are looping over the items in the set , and print each one without new line at the end. In the first code, the condition will return True as long as *args is not None .

def function(*args):
    if *args:
        return f'{args} is not None!`

Answer 4

The first example prints a newline character more if args contains one ore more arguments. I think it's just there to make the output a bit prettier and it is not really important.