The same object is modified in the function.
Let’s break this down step by step so that it’s clear.
Code Analysis:
The Function:
def foo(x):
x[0] = [‘def’] # Changes the first element of the list `x` to [‘def’]
x[1] = [‘abc’] # Changes the second element of the list `x` to [‘abc’]
return id(x) # Returns the memory address (ID) of the modified list `x`
The Main Code:
q = [‘abc’, ‘def’] # Creates a list `q` with two elements: [‘abc’, ‘def’]
print(id(q) == foo(q)) # Compares the memory address (ID) of `q` with the value returned by `foo(q)`
Step-by-Step Explanation:
1. List q Initialization:
• A list q is created:
q = [‘abc’, ‘def’]
• This list has an ID, which is its memory address. Let’s call it id(q).
2. Calling foo(q):
• The list q is passed into the function foo.
• Inside foo:
• x[0] = [‘def’]: The first element of the list is changed to [‘def’].
Now, q looks like this:
[[‘def’], ‘def’]
• x[1] = [‘abc’]: The second element of the list is changed to [‘abc’].
Now, q looks like this:
[[‘def’], [‘abc’]]
• The id(x) is returned. Since x and q refer to the same list, id(x) is the same as id(q).
3. Comparison (id(q) == foo(q)):
• The memory address of q is compared to the memory address returned by foo(q).
• Since foo(q) does not create a new list but modifies the same list, id(q) and id(x) are identical.
Output:
True
Key Points to Understand:
1. Lists are mutable: When you pass a list to a function, any changes made to the list inside the function will affect the original list because they share the same memory address.
2. The id() function: This returns the memory address of an object.
The same object is modified in the function.
Let’s break this down step by step so that it’s clear.
Code Analysis:
The Function:
def foo(x):
x[0] = [‘def’] # Changes the first element of the list `x` to [‘def’]
x[1] = [‘abc’] # Changes the second element of the list `x` to [‘abc’]
return id(x) # Returns the memory address (ID) of the modified list `x`
The Main Code:
q = [‘abc’, ‘def’] # Creates a list `q` with two elements: [‘abc’, ‘def’]
print(id(q) == foo(q)) # Compares the memory address (ID) of `q` with the value returned by `foo(q)`
Step-by-Step Explanation:
1. List q Initialization:
• A list q is created:
q = [‘abc’, ‘def’]
• This list has an ID, which is its memory address. Let’s call it id(q).
2. Calling foo(q):
• The list q is passed into the function foo.
• Inside foo:
• x[0] = [‘def’]: The first element of the list is changed to [‘def’].
Now, q looks like this:
[[‘def’], ‘def’]
• x[1] = [‘abc’]: The second element of the list is changed to [‘abc’].
Now, q looks like this:
[[‘def’], [‘abc’]]
• The id(x) is returned. Since x and q refer to the same list, id(x) is the same as id(q).
3. Comparison (id(q) == foo(q)):
• The memory address of q is compared to the memory address returned by foo(q).
• Since foo(q) does not create a new list but modifies the same list, id(q) and id(x) are identical.
Output:
True
Key Points to Understand:
1. Lists are mutable: When you pass a list to a function, any changes made to the list inside the function will affect the original list because they share the same memory address.
2. The id() function: This returns the memory address of an object.