References, (Im)mutability

Immutability: Numbers

Numbers are immutable …

• Object of type int with value 42

• Variable a points to it (“gives it a name”)

• The object cannot change its value - there is no method to modify an integer object

• ⟶ The latter situation is equivalent to the former (which is the implementation)

a = 42 b = a Both a and b refer to the same object: print(id(a)) print(id(b)) 140418542006832 140418542006832

Modifying An Integer In-Place? (Immutability)

id(a)

140418542006832

• Operator += modifies an integer in-place?

• a = a+7 modifies an integer in-place?

• ⟶ No, both create new objects!

a += 1
id(a)         # <--- different object now

140418542006864

a = a + 7
id(a)         # <--- again, different object now

140418542007088

Immutability: Tuples

a = (42, "xy", 13) b = a print(id(a)) print(id(b)) 140418186318912 140418186318912 Like lists, but immutable No way to modify a tuple No appending No slice assignment No nothing

• And operator +=?

a += ('foo', 'bar')
print(a)
print(id(a))
print(id(b))

(42, 'xy', 13, 'foo', 'bar')
140418252413568
140418186318912

• ⟶ leaves b alone

b

(42, 'xy', 13)

Mutability: Lists (1)

Lists are mutable …

• Objects can be modified

• E.g. by using append()

a = [1, 2, 3] b = a b [1, 2, 3] Both now refer to the same object Modify b … b.append(4) b [1, 2, 3, 4] … and see a modified! a [1, 2, 3, 4]

Mutability: Lists (2)

Danger …

• Take care when passing complex data structures

• Not passed by copy (as in C++)

• Passed by reference (as in Java and C#, for example)

Solution?

• Is copy a solution?

• ⟶ I’d say no!

• Being careful is a solution!

a = [1, 2, 3]
b = a[:]             # <--- copy
a.append(4)
b

[1, 2, 3]

Shallow Copy

• A list within a list

• Create “copy”

a = [1, [1, 2, 3], 2]
b = a[:]      # <--- copy
b

[1, [1, 2, 3], 2]

• This is only a shallow copy

• Modify a

a[1].append(4)
a

[1, [1, 2, 3, 4], 2]

• And b?

b

[1, [1, 2, 3, 4], 2]

• Reason: nested list has not been copied!

a[1] is b[1]

True

• Only first level copied

• Shallow copy

• a[1] is a reference

• is: object identity

Deep Copy

Solution: not easy

• Recursive structure traversal

• Handling every possible type

• Dedicated module in the standard library: copy

• Solution?

import copy
a = [1, [1, 2, 3], 2]
b = copy.deepcopy(a)
a[1] is b[1]

False