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

Question

What will be the output of the following Python code?

def foo(k):
    k = [1]

q = [0]
foo(q)
print(q)

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Solution

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

Here's the step by step explanation:

A list q is defined with one element: 0.
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 to functions by reference, and the function only changes the local reference k, not q.
After the function call, print(q) is executed. Since q has not been modified by the function foo(k), it remains as [0].

So, the output of the code is [0].

This problem has been solved

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