[Solved] Which of the following is the correct way to declare a mutable list interface in Kotlin? — Ans: val list: MutableList = mutableListOf(1, 2, 3) | Kotlin

Kotlin - Collections Fundamentals

Which of the following is the correct way to declare a mutable list interface in Kotlin?

Aval list: MutableList<Int> = listOf(1, 2, 3)

Bval list: List<Int> = mutableListOf(1, 2, 3)

Cval list: MutableList<Int> = mutableListOf(1, 2, 3)

Dval list: List<Int> = listOf(1, 2, 3)

Step-by-Step Solution

Solution:

Step 1: Identify mutable list declaration
MutableList is the interface for mutable lists, and mutableListOf() creates a mutable list instance.
Step 2: Check each option
val list: MutableList = mutableListOf(1, 2, 3) correctly declares a MutableList and initializes it with mutableListOf().
Final Answer:
val list: MutableList = mutableListOf(1, 2, 3) -> Option C
Quick Check:
MutableList + mutableListOf() = mutable list [OK]

Quick Trick: Use MutableList with mutableListOf() for mutable lists [OK]

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Which of the following is the correct way to declare a mutable list interface in Kotlin?