CODEWITHSUNDEEP
python
ealeon
ealeon

Reputation: 12452

Python raw_input()

From the docs, raw_input() reads a line from input, converts it to a string (stripping a trailing newline), and returns that.

with that note,

a = 'testing: '
sep = '-'
c = raw_input('give me some args: ')          <--- giving 'a b c d'

def wrap( a, sep, *c):
    print a + sep.join(c)

wrap(a, sep, c)

str = 'a b c d'
print sep.join(str)

they should both print out the same thing but...

print a + sep.join(c) gives testing: a b c d
print sep.join(str) gives a- -b- -c- -d

why doesn't sep.join() works inside wrap function?

EDIT changing from *c to c makes the output the same but this somewhat confuses me because i thought *c unpacks the args but when i print c, it give ms ('a b c d',) compared to a string of 'a b c d' so in a sense, it is combining them to a single tuple entity which is the opposite of unpacking?

or... it does not unpack string but only lists?

Upvotes: 1

Views: 607

Answers (3)

dnozay
dnozay

Reputation: 24304

this:

>>> wrap(a, sep, str)
testing: a b c d

is different from:

>>> wrap(a, sep, *str)
testing: a- -b- -c- -d

A string is a list of characters. You can change the function signature from def wrap(a, sep, *c): to def wrap(a, sep, c): and you should be fine.

Upvotes: 1

Steve Barnes
Steve Barnes

Reputation: 28360

join expects an array not a string so if you are trying to use the arguments as separate items you need to use c = c.spilt(' ') and get rid of the * in the def wrap.

To my surprise sep.join(str) is treating str as an array of chars hence the extra -s between letters and spaces.

Upvotes: 2

Dek Dekku
Dek Dekku

Reputation: 1461

In your function c is a tuple of one element because of the splat (*), therefore the separator is never used. In the main program, you're calling join on a string, which is split into its characters. Then you get as many istances of the separator as many charactes are there (minus one).

Upvotes: 3

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