On Tue, 25 Dec 2012 18:00:38 -0800, Abhas Bhattacharya wrote:
> While I am defining a function, how can I access the name (separately as
> string as well as object) of the function without explicitly naming
> it(hard-coding the name)? For eg. I am writing like:
> def abc():
> #how do i access the function abc here without hard-coding the name?
You can't easily get the function name from inside the function. But you
*can* easily get the function name from outside the function, so the
simple solution is to use a decorator to wrap the function with something
that knows its name. For example:
import functools
def print_function_name(func):
# Decorate func so that it prints a message when called.
@functools.wraps(func)
def inner(*args, **kwargs):
print "Calling function %s" % func.__name__
return func(*args, **kwargs)
return inner
@print_function_name
def spam(n):
return ' '.join(['spam']*n)
@print_function_name
def breakfast(n):
return ' and '.join(['ham', 'eggs', spam(n)])
And then in use:
py> spam(3)
Calling function spam
'spam spam spam'
py>
py> breakfast(5)
Calling function breakfast
Calling function spam
'ham and eggs and spam spam spam spam spam'
I recommend you use this solution if possible.
Unfortunately this doesn't help if you actually need the function name
*inside* the function, for example:
def invert(n):
if n != 0:
return 1/n
else:
raise ZeroDivisionError(
'divide by zero error in function %s' % ***MAGIC GOES HERE***)
(But note, why would I bother doing this? When the traceback prints, it
already displays the function name. Why am I doing by hand what Python
automatically does for me?)
If you really must do this, you can do it like this:
import traceback
def get_current_function_name():
"""If this looks like magic, that's because it is."""
x = traceback.extract_stack(limit=2)
return x[0][2]
def invert(n):
if n != 0:
return 1/n
else:
me = get_current_function_name()
raise ZeroDivisionError('divide by zero error in func %s' % me)
--
Steven
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