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@Python_Dv B) [1] a = b = [] → both point to the SAME list a.append(1) → modifies that shared list print(b) → [1] The tricky part: "=" doesn't copy, it creates a reference. Welcome to Python's mutable object trap.

@Python_Dv Output: B) [1] Explanation: a and b point to the same list. a.append(1) mutates that shared list, so b sees the change too.

@Python_Dv Option B : [1]

@Python_Dv B) [1] a and b reference the same list, so appending via a affects b too.

@Python_Dv (B) [1] When the line a = b = [] is executed in Python, a single empty list object is created in memory, and both variables a and b are made to refer to this exact same object. Lists are mutable objects, meaning their content can be changed after creation without creating a new

@Python_Dv B because a gets appended and a=b so B get’s what A gets.

@Python_Dv B. `a` and `b` refer to the same object.

@Python_Dv B

@Python_Dv answer- []

@Python_Dv B

@Python_Dv @grok What is the anwser, my guess is B

@Python_Dv C

@Python_Dv a = b = [] → both a and b refer to the same list But oppened is not a valid list method So Python raises an Error

@Python_Dv B

@Python_Dv B)

@Python_Dv B) [1]

@Python_Dv B

@Python_Dv Error

@Python_Dv B) [1]

@Python_Dv B

@Python_Dv Error It’s append, not oppend

@Python_Dv 1

@Python_Dv Error