def f(x,l=[]): for i in range(x): l.append(i*i) print(l) f(2) f(3,[3,2,1]) f(3)

The output of the code is:

[0, 1]
[3, 2, 1, 0, 1, 4]
[0, 1, 0, 1, 4]

Explanation:
- The function `f()` takes two parameters: `x` and `l` (which is a list with a default value of an empty list `[]`).
- Inside the function, a loop runs from 0 to `x-1`, and for each iteration, the square of `i` is appended to the list `l`.
- Finally, the function prints the contents of `l`.
- When `f(2)` is called, the loop runs twice, and the squares of 0 and 1 are added to the list `l`, which becomes `[0, 1]`.
- When `f(3, [3, 2, 1])` is called, the list `[3, 2, 1]` is passed as the value of `l`, so the loop runs thrice, and the squares of 0, 1, and 2 are added to the list `l` (which already has the values `[3, 2, 1]`), so it becomes `[3, 2, 1, 0, 1, 4]`.
- When `f(3)` is called again (without passing any value for `l`), the default list `[]` is used, which already has the values `[0, 1, 0, 1, 4]` from the previous call, so the loop runs thrice again, and the squares of 0, 1, and 2 are added to the list `l`, which becomes `[0, 1, 0, 1, 4, 0, 1, 4]`.