The output of the given Python code will be [0].

Here's the step by step explanation:

1. A list `q` is defined with a single element 0.

2. The function `foo(k)` is called with `q` as the argument. Inside the function, `k` is assigned a new list [1]. However, this doesn't affect `q` because lists are passed by reference in Python, and the function is creating a new reference `k` instead of modifying the original reference `q`.

3. After the function call, `q` is printed, which is still [0] as it was not modified by the function.

Question

The output of the given Python code will be [0].

Here's the step by step explanation:

1. A list `q` is defined with a single element 0.

2. The function `foo(k)` is called with `q` as the argument. Inside the function, `k` is assigned a new list [1]. However, this doesn't affect `q` because lists are passed by reference in Python, and the function is creating a new reference `k` instead of modifying the original reference `q`.

3. After the function call, `q` is printed, which is still [0] as it was not modified by the function.

Knowee AI · Accepted Answer

The output of the given Python code will be [0].

Here's the step by step explanation:

1. A list `q` is defined with a single element 0.

2. The function `foo(k)` is called with `q` as the argument. Inside the function, `k` is assigned a new list [1]. However, this doesn't affect `q` because lists are passed by reference in Python, and the function is creating a new reference `k` instead of modifying the original reference `q`.

3. After the function call, `q` is printed, which is still [0] as it was not modified by the function.

What will be the output of the following Python code?def foo(k): k = [1]q = [0]foo(q)print(q) [0][1][1, 0][0, 1]

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What will be the output of the following Python code?

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