> when i call a method foo from another method func. can i access func context
> variables or locals() from foo
> so
> def func():
> i=10
> foo()
>
> in foo, can i access func's local variables
A. python has statically-nested scoping, so you can do it as long as you:
1. define foo() as an inner function -- its def is contained within
func()'s def:
def func():
i = 10
def foo():
print i
2. you don't define a variable with the same name locally. in other
words, you do have access to func()'s 'i' as long as you don't create
another 'i' within foo() -- if you do, only your new local 'i' will be
within your scope.
B. another alterative is to pass in all of func()'s local variables to foo():
foo(**locals())
this will require you to accept the incoming dictionary in foo() and
access the variables from it instead of having them be in foo()'s
scope.
C. in a related note, your question is similar to that of global vs.
local variables. inside a function, you have access to the global as
long as you don't define a local using the same name. should you only
wish to manipulate the global one, i.e. assign a new value to it
instead of creating a new local variable with that name, you would
need to use the "global" keyword to specify that you only desire to
use and update the global one.
i = 0
def func():
i = 10
in this example, the local 'i' in func() hides access to the global
'i'. in the next code snippet, you state you only want to
access/update the global 'i'... IOW, don't create a local one:
i = 0
def func():
global i
i = 10
D. this doesn't work well for inner functions yet:
i = 0
def func():
i = 10
def foo():
i = 20
the 'i' in foo() shadows/hides access to func()'s 'i' as well as the
global 'i'. if you issue the 'global' keyword, that only gives you
access to the global one:
i = 0
def func():
i = 10
def foo():
global i
i = 20
you cannot get access to func()'s 'i' in this case.
E. however, starting in Python 3.x, you'll be able to do somewhat
better with the new 'nonlocal' keyword:
i = 0
print('globally, i ==', i)
def func():
i = 10
print('in func(), i ==', i)
def foo():
nonlocal i
i = 20
print('in foo(), i ==', i)
foo()
print('in func() after calling foo(), i ==', i)
func()
in this case, foo() modified func()'s 'i'.
hope this helps!
-- wesley
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
"Core Python Programming", Prentice Hall, (c)2007,2001
"Python Fundamentals", Prentice Hall, (c)2009
http://corepython.com
wesley.j.chun :: wescpy-at-gmail.com
python training and technical consulting
cyberweb.consulting : silicon valley, ca
http://cyberwebconsulting.com
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