Why is Java secure?

Why is Java secure?

• A. It relies on OS security features only
• B. It does not use pointers and prevents out-of-bounds array access
• C. It uses manual memory management
• D. It allows direct memory access via pointers

Answer: (B)

Java’s security comes from memory safety enforced by the language and its runtime. You don’t work with raw pointers; references can’t be used for pointer arithmetic or to access arbitrary memory. Each array access is checked at runtime, so attempts to read or write outside bounds are blocked with an exception instead of corrupting memory. Automatic memory management through garbage collection removes many risks tied to manual memory handling, such as freeing or reusing memory incorrectly. While OS protections help, the built-in safety features—no pointers and bounds-checked memory access—provide the strongest, language-level security guarantees.

Java’s security comes from memory safety enforced by the language and its runtime. You don’t work with raw pointers; references can’t be used for pointer arithmetic or to access arbitrary memory. Each array access is checked at runtime, so attempts to read or write outside bounds are blocked with an exception instead of corrupting memory. Automatic memory management through garbage collection removes many risks tied to manual memory handling, such as freeing or reusing memory incorrectly. While OS protections help, the built-in safety features—no pointers and bounds-checked memory access—provide the strongest, language-level security guarantees.