CODEWITHSUNDEEP
pythonobjectmemory-managementimmutabilitycpu-architecture
Chitranjan
Chitranjan

Reputation: 67

How does memory location changes by modifying an immutable object in python?

We can check if the object type is mutable or not by comparing the new memory location of modified variable with the originally defined variable.

For example - int() is an immutable object in Python. If I try to modify the value of integer type variable, I notice the memory location changes [Code and Output below]. Can someone provide a brief explanation going in the background ?

#initial variable
a = 10
# initial memory location
print(id(a))
#modified variable
a += 1
# new memory location, is it same?
print(id(a))

OUTPUT

93285870446416
93285870446524

Upvotes: 3

Views: 1555

Answers (2)

Luc Bertin
Luc Bertin

Reputation: 336

Note that contrarily to C, and static-typing languages, Python type-checks only at runtime.

in C you would do:

int a; /* creating a memory space allowing only holding data of integer type */
a = 3; /* storing the value 3, not a character, but an integer, otherwise error is returned at compilation */

in Python, a = 3 first creates an object 3 at a certain memory space, and then links the name a to that object location during assignment. a is then bound to that object by pointing to its memory allocated.

CPython already allocates a memory space for integers in range -5 to 256. If you write now b = 3, b will share the same memory location as b is simply another pointer to the same object 3.

Integers being immutable, if you do a += 1, then a new object 4 is created at a different memory address and then a will points to this new location. b stays bound to the object 3.

Note that if an object is not linked anymore to any names (a, b, etc), then it can be garbage-collected. A counter sys.getrefcount(X) is used to keep track of all the references to a given object "X".

Upvotes: 1

Thomas Weller
Thomas Weller

Reputation: 59238

What's going on in the background?

a = 10
  1. The interpreter sees an expression and determines the type of the result, which is int
  2. Memory for an int is reserved, giving a memory address, marking that memory as "in use"
  3. The result of the operation is stored at that memory address
  4. -/- (not applicable yet)
a += 1
  1. The interpreter sees an expression and determines the type of the result, which is int
  2. Memory for an int is reserved, giving a new memory address, because the old one is still in use
  3. The result of the operation is stored at that new memory address
  4. There is no variable poining to the old address any more, so that one is marked as free

Maybe another

a += 1
  1. The interpreter sees an expression and determines the type of the result, which is int
  2. Memory for an int is reserved, giving a either another new memory address, or reuse the now freed memory address ("old" address)
  3. The result of the operation is stored at that new memory address
  4. There is no variable poining to the address any more, so that one is marked as free

and so on

Upvotes: 7

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