Why this way of using fucntions arguments does not work?

Question

Why does this function print arg1 arg2 instead of Hello world? Shouldn't arg be equal to the arguments arg1 and arg2 and then be used in print()?

def printer(arg1, arg2):
    for i in range(2):
        arg = 'arg{} '.format(str(i+1))
        print(arg, end="")

printer('Hello', 'world')

A string containing the name of a variable is not the variable. It "doesn't work" because it's wrong. That is to say, it works fine; it does exactly what you told it to do. — kindall, Feb 05 '18 at 03:08
It's not at all clear that this is an appropriate duplicate. OP is clearly interested in why Python doesn't recognize strings as symbols. — Joshua R., Feb 05 '18 at 03:23

score 1 · Answer 1 · answered Feb 05 '18 at 03:14

You are trying to print strings, not your arguments. If you need to iterate through your function arguments, you can do something like this:

Code:

def printer(*args):
    for arg in args:
        print(arg, end="")

printer('Hello', 'world')

Results:

Helloworld

score 1 · Accepted Answer · answered Feb 05 '18 at 03:19

1

You can use locals() to access current local scope:

def printer(arg1, arg2):
    for i in range(2):
        arg = 'arg{}'.format(str(i+1))
        print(locals().get(arg), end="")

printer('Hello', 'world')

answered Feb 05 '18 at 03:19

Sraw

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score -1 · Answer 3 · answered Feb 05 '18 at 03:16

@kindle is right. You're making a string variable. And Stephen Rauch's answer is the a better way to accomplish what you want.

However, Python can be made to evaluate a string as code using the eval function. You could get the effect you expected by changing

print(arg, end="")

to

print(eval(arg), end="")

Why this way of using fucntions arguments does not work?

3 Answers3

Code:

Results: