Python: some newbie questions on sys.stderr and using function as argument
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by Cawas
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Published on 2010-04-14T00:16:35Z
Indexed on
2010/04/14
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I'm just starting on Python and maybe I'm worrying too much too soon, but anyways...
log = "/tmp/trefnoc.log"
def logThis (text, display=""):
msg = str(now.strftime("%Y-%m-%d %H:%M")) + " TREfNOC: " + text
if display != None:
print msg + display
logfile = open(log, "a")
logfile.write(msg + "\n")
logfile.close()
return msg
def logThisAndExit (text, display=""):
msg = logThis(text, display=None)
sys.exit(msg + display)
That is working, but I don't like how it looks. Is there a better way to write this (maybe with just 1 function) and is there any other thing I should be concerned under exiting?
Now to some background...
Sometimes I will call logThis
just to log and display. Other times I want to call it and exit. Initially I was doing this:
logThis ("ERROR. EXITING")
sys.exit()
Then I figured that wouldn't properly set the stderr
, thus the current code shown on the top.
My first idea was actually passing "sys.exit" as an argument, and defining just logThis ("ERROR. EXITING", call=sys.exit)
defined as following (showing just the relevant differenced part):
def logThis (text, display="", call=print):
msg = str(now.strftime("%Y-%m-%d %H:%M")) + " TREfNOC: " + text
call msg + display
But that obviously didn't work. I think Python doesn't store functions inside variables. I couldn't (quickly) find anywhere if Python can have variables taking functions or not! Maybe using an eval
function? I really always try to avoid them, tho. Sure I thought of using if
instead of another def
, but that wouldn't be any better or worst.
Anyway, any thoughts?
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