Python : When is a variable passed by reference and when by value?


Possible Duplicate:
Python: How do I pass a variable by reference?

My code :

locs = [ [1], [2] ]
for loc in locs:
    loc = []

print locs
# prints => [ [1], [2] ]

Why is loc not reference of elements of locs ?

Python : Everything is passed as reference unless explicitly copied [ Is this not True ? ]

Please explain.. how does python decides referencing and copying ?

Update :

How to do ?

def compute(ob):
   if isinstance(ob,list): return process_list(ob)
   if isinstance(ob,dict): return process_dict(ob)

for loc in locs:
   loc = compute(loc)  # What to change here to make loc a reference of actual locs iteration ?
  • locs must contain the final processed response !
  • I don't want to use enumerate, is it possible without it ?

Best Solution

Effbot (aka Fredrik Lundh) has described Python's variable passing style as call-by-object:

Objects are allocated on the heap and pointers to them can be passed around anywhere.

  • When you make an assignment such as x = 1000, a dictionary entry is created that maps the string "x" in the current namespace to a pointer to the integer object containing one thousand.
  • When you update "x" with x = 2000, a new integer object is created and the dictionary is updated to point at the new object. The old one thousand object is unchanged (and may or may not be alive depending on whether anything else refers to the object).
  • When you do a new assignment such as y = x, a new dictionary entry "y" is created that points to the same object as the entry for "x".
  • Objects like strings and integers are immutable. This simply means that there are no methods that can change the object after it has been created. For example, once the integer object one-thousand is created, it will never change. Math is done by creating new integer objects.
  • Objects like lists are mutable. This means that the contents of the object can be changed by anything pointing to the object. For example, x = []; y = x; x.append(10); print y will print [10]. The empty list was created. Both "x" and "y" point to the same list. The append method mutates (updates) the list object (like adding a record to a database) and the result is visible to both "x" and "y" (just as a database update would be visible to every connection to that database).

Hope that clarifies the issue for you.