[Solved] Why does the safe call operator ?. not prevent exceptions like IndexOutOfBoundsException in Kotlin? — Ans: Because it only checks for null, not invalid indices or other errors | Kotlin

Kotlin - Null Safety

Why does the safe call operator ?. not prevent exceptions like IndexOutOfBoundsException in Kotlin?

ABecause it converts exceptions to null silently

BBecause it disables all exceptions including index errors

CBecause it only checks for null, not invalid indices or other errors

DBecause it only works with primitive types

Step-by-Step Solution

Solution:

Step 1: Understand safe call operator purpose
The safe call operator only checks if the object is null before calling a method or property.
Step 2: Recognize limitations of safe call
It does not handle exceptions caused by invalid operations like accessing invalid list indices.
Final Answer:
Because it only checks for null, not invalid indices or other errors -> Option C
Quick Check:
Safe call protects null, not other runtime exceptions [OK]

Quick Trick: Safe call checks null only, not other exceptions [OK]

Common Mistakes:

MISTAKES

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Why does the safe call operator ?. not prevent exceptions like IndexOutOfBoundsException in Kotlin?