[Tutor] exception about ctrl+c

2012-01-09 Thread daedae11 
I want to catch the ctrl+c exception. My program is as following. But when I 
run my script and press ctrl+c, the  program output nothing. I don't know 
where did I go wrong. Please help me. Thank you!

def safe_input(prompting):
try:
return raw_input(prompting);
except KeyboardInterrupt, error:
print error;
return None;

def main():
a = safe_input(input any thing!\n);
print a;


if __name__ == '__main__':
main();




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Re: [Tutor] exception about ctrl+c

2012-01-09 Thread Joel Goldstick 
On Mon, Jan 9, 2012 at 7:24 AM, daedae11 daeda...@126.com wrote:
 I want to catch the ctrl+c exception. My program is as following. But when
 I run my script and press ctrl+c, the  program output nothing. I don't
 know where did I go wrong. Please help me. Thank you!

 def safe_input(prompting):
 try:
 return raw_input(prompting);
 except KeyboardInterrupt, error:
 print error;
 return None;

 def main():
 a = safe_input(input any thing!\n);
 print a;


 if __name__ == '__main__':
 main();

 
 daedae11

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  python ctl-c.py
  input any thing!
  fdsaj
  fdsaj
  python ctl-c.py
  input any thing!
  ^C
  None

I just got this from your code.  Seems to work on python 2.65, linux

Are you running it from a command line like:  python ctl-c.py
or are you running in a python shell?  If you are in a shell it might
be consuming the ctl-c before your program can


-- 
Joel Goldstick
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Re: [Tutor] exception about ctrl+c

2012-01-09 Thread Christian Witts 

On 2012/01/09 02:24 PM, daedae11 wrote:
I want to catch the ctrl+c exception. My program is as following. 
But when I run my script and press ctrl+c, the  program output 
nothing. I don't know where did I go wrong. Please help me. Thank you!

def safe_input(prompting):
try:
return raw_input(prompting);
except KeyboardInterrupt, error:
print error;
return None;
def main():
a = safe_input(input any thing!\n);
print a;
if __name__ == '__main__':
main();

daedae11


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def safe_input(prompting):
try:
return raw_input(prompting)
except KeyboardInterrupt:
print 'KeyboardInterrupt Issued'
return None

That will work as intended, if you had your original `except 
KeyboardInterrupt, error:` and did a `print repr(error)` afterwards you 
will see it does not contain an error message as you perhaps wanted.


Also, Python does not require semi-colons to denote the end-of-line. It 
can be used if you want to have multiple statements on a single line though.


--

Christian Witts
Python Developer
//
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