this is written in python 3.3 and i''m getting error as syntax error. Can any one fix this?

Question

while True:
    n=int(raw_input())
    if n!=42:
        print n
    else:
        break

Can you show us the traceback of the error? (ie what the interpreter prints out when it tells you you have a syntax error) That will help us help you figure your problem out. — Sam Mussmann, Dec 02 '12 at 00:09
Use code formatting and line breaks so we can actually see your syntax. — pascalhein, Dec 02 '12 at 00:10

score 3 · Answer 1 · answered Dec 02 '12 at 01:37

3

Get a Python 3.X tutorial. Python 3.X introduces non-backward-compatible changes to the language. raw_input no longer exists and print is a function instead of a statement in Python 3.X:

Corrected code:

while True:
    n=int(input())
    if n!=42:
        print(n)
    else:
        break

answered Dec 02 '12 at 01:37

Mark Tolonen

166,664
26
169
251

Thank you for correcting both errors! – Noctis Skytower Dec 02 '12 at 02:37

unutbu · Answer 2 · 2012-12-02T01:57:41.003

1

print is a function (not a statement) in Python3. Use

print(n)

instead of

print n

edited Dec 02 '12 at 01:57

answered Dec 02 '12 at 00:12

unutbu

842,883
184
1,785
1,677

That is a problem - but I would have thought the OP to have hit a `NameError` with `raw_input` first :) – Jon Clements Dec 02 '12 at 01:37

score 1 · Answer 3 · answered Dec 02 '12 at 00:12

1

Use parentheses around the parameter of a function: print(n) instead of print n

answered Dec 02 '12 at 00:12

pascalhein

5,700
4
31
44

this is written in python 3.3 and i''m getting error as syntax error. Can any one fix this?

3 Answers3