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 ?
enumerate
, is it possible without it ?Python always uses pass-by-reference values. There isn't any exception. Any variable assignment means copying the reference value.
Pass by value refers to a mechanism of copying the function parameter value to another variable while the pass by reference refers to a mechanism of passing the actual parameters to the function. Thus, this is the main difference between pass by value and pass by reference.
Python utilizes a system, which is known as “Call by Object Reference” or “Call by assignment”. In the event that you pass arguments like whole numbers, strings or tuples to a function, the passing is like call-by-value because you can not change the value of the immutable objects being passed to the function.
Passing by reference means the called functions' parameter will be the same as the callers' passed argument (not the value, but the identity - the variable itself). Pass by value means the called functions' parameter will be a copy of the callers' passed argument.
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.
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. 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).y = x
, a new dictionary entry "y" is created that points to the same object as the entry for "x".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.
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.
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