When is a variable passed by reference and when by value? [duplicate]

Question

My code:

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

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

Why is loc not reference of elements of locs?

In Python, everything is passed as reference unless explicitly copied (is this not true?)

How does Python decide referencing and copying?

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?

Answer 1

Effbot (aka Fredrik Lundh) has described Python's variable passing style as call-by-object: http://effbot.org/zone/call-by-object.htm

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.

Answer 2

Everything in Python is passed and assigned by value, in the same way that everything is passed and assigned by value in Java. Every value in Python is a reference (pointer) to an object. Objects cannot be values. Assignment always copies the value (which is a pointer); two such pointers can thus point to the same object. Objects are never copied unless you're doing something explicit to copy them.

For your case, every iteration of the loop assigns an element of the list into the variable loc. You then assign something else to the variable loc. All these values are pointers; you're assigning pointers; but you do not affect any objects in any way.

Answer 3

It doesn't help in Python to think in terms of references or values. Neither is correct.

In Python, variables are just names. In your for loop, loc is just a name that points to the current element in the list. Doing loc = [] simply rebinds the name loc to a different list, leaving the original version alone.

But since in your example, each element is a list, you could actually mutate that element, and that would be reflected in the original list:

for loc in locs:
    loc[0] = loc[0] * 2

Answer 4

When you say

loc = []

you are rebinding the loc variable to a newly created empty list

Perhaps you want

loc[:] = []

Which assigns a slice (which happens to be the whole list) of loc to the empty list

Answer 5

Everything is passed by object. Rebinding and mutating are different operations.

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

print locs

Answer 6

Why is loc not reference of elements of locs ?

It is. Or at least, it is in the same sense that every other variable in Python is. Python variables are names, not storage. loc is a name that is used to refer to elements of [[1,2], [3,4]], while locs is a name that refers to the entire structure.

loc = []

This does not mean "look at the thing that loc names, and cause it to turn into []". It cannot mean that, because Python objects are not capable of such a thing.

Instead, it means "cause loc to stop being a name for the thing that it's currently a name for, and start instead being a name for []". (Of course, it means the specific [] that's provided there, since in general there may be several objects in memory that are the same.)

Naturally, the contents of locs are unchanged as a result.