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Option C : [1, 2, 3] [1, 2, 3, 4]
Explanation👇
a = [1,2,3] creates a list.
b = a[:] creates a copy of list `a` using slicing. Now `b` is a **separate list in memory.
b.append(4) adds `4` only to `b`, not to `a`.
Since `a` and `b` are different lists, modifying `b` does not affect `a`.
Key point:
`a[:]` makes a **shallow copy**, not a reference.
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@Python_Dv Answer: (C)
This problem tests the understanding of list copying in Python.
Variable Assignment: The line a = [1, 2, 3] creates a list and assigns it to variable a.
Slicing for Copy: The expression a[:] creates a shallow copy of the list a and assigns it to b. This
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@Python_Dv It is c because a is not appended it was b that was appended.
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@Python_Dv Answer: C. Slice [:] copies the list → a stays [1,2,3], b becomes [1,2,3,4].
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@Python_Dv output:
👉 [1, 2, 3] [1, 2, 3, 4]
Reason?
a[:] creates a shallow copy of the list.
So b is a new list.
Appending to b doesn’t affect a.
b = a
Then both would change. 😄
Copy vs reference.
Classic Python interview trap.
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C. [1, 2, 3] [1, 2, 3, 4].
Two concepts are in play here. The first is the concept of slicing and the second one is the append method.
b = a[ : ]. This means slice through "a" from the beginning to the end. Which gives [1, 2, 3].
b.append(4). Means add 4 to the existing values of b, but as the last element. This gives [1, 2, 3, 4].
So print(a, b) gives the answer.
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@Python_Dv 1. b = a[:] this type of assignment is an old fashioned way of copying. This means variable b is an separate memory from variable a. Changes made to variable b won't affect variable a
2. Output will be [1,2,3][1,2,3,4]
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@Python_Dv C. We have two lists in differtmemory locations. Only b is modified in place. So the result is[1,2,3] [1,2,3,4]
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@Python_Dv b = a[:] creates a 'shallow copy' of a. Now a and b share different memory and .append() outputs option 'C'.
[:] is a slicing technique which copies everything from start to end. Unlike append(), slicing creates a brand new list without altering original list.
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