Functions and objects

[Mark Hathaway]

>> Matthew wrote:
>>
>> I'm confused.  Take a look at the following:

a = [5]

>> for x in range(5):     # Case A
>>     temp=a
>>     print a
>>     temp.append(1)
       print temp        # to see what's happening to it

>> [5]
>> [5, 1]
>> [5, 1, 1]
>> [5, 1, 1, 1]
>> [5, 1, 1, 1, 1]

a = [5]

>> for x in range(5):     # Case B
>>     temp=a[:]
>>     print a
>>     temp.append(1)
       print temp        # to see what's happening to it

>> [5]
>> [5]
>> [5]
>> [5]
>> [5]


>> In case A, why doesn't temp reset itself to the value of a, [5], that
>> was predetermined before it entered the loop?  Why do you need to copy
>> the list to get the behavior I'm looking for in Case B?  Doesn't a
>> always equal [5]?

> Assignment always stores a reference to an object, it DOES NOT make a
> copy of the object.

It doesn't default to making a copy. you have to specify if you want a
copy!

So,

    temp = a

isn't the same as

    temp = a[:]

The first copies the address, stored as a simple variable in a, which
points to the object [5]

The second doesn't copy the address, it creates a new object with a copy
of what a pointed to.

This is one of the oddities (IMO) of Python.

It's about like saying temp = a[:] is temp = newOjbect([5])
wheras                 temp = a    is temp = addressOf([5])

I know it was all simple when you could write (in most languages)

    a = 5
    temp = a

and temp is a new separate different variable with a value of 5.
That's not quite what's happening in Python. Study the difference
carefully.


Mark S. Hathaway
email: hathawa2 at marshall.edu