[EMAIL PROTECTED] wrote:
> ... the first element of the list to which x refers is a reference to
> the new string and back outside foo, the first element of the list to
> which x refers will be a reference to the new string.
I'd rephrase that as:
* Both the global context and the inside of foo see the same list
* They can therefore both update the list
* If a new string is put in the first element of the list, the can
both see the same new string.
> Right?
You know you can get python to answer your question - yes? Might be slightly
more illuminating than twisting round english... :-)
OK, you're passing in a string in a list. You have 2 obvious ways of doing
that - either as an argument:
def foo(y):
y[0] += " other"
print id(y[0])
... or as a global: (which of course you wouldn't do :-)
def bar():
global x
x[0] += " another"
print id(x[0])
So let's see what happens.
>>> x = ["some string"] # create container with string
>>> x[0] # Check that looks good & it does
'some string'
>>> id(x[0]) # What's the id of that string??
3082578144L
>>> foo(x) # OK, foo thinks the new string has the following id
3082534160
>>> x[0] # Yep, our x[0] has updated, as expected.
'some string other'
>>> id(x[0]) # Not only that the string has the same id.
3082534160L
>>> bar() # Update the global var, next line is new id
3082543416
>>> x[0] # Check the value's updated as expected
'some string other another'
>>> id(x[0]) # Note that the id is the same as the output from bar
3082543416L
Does that perhaps answer your question more precisely ?
Michael.
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