Given the code fragment: List<String> items = new ArrayList<> (); items.add(1, "pen"); items.add(2, "pencil"); items.add(3, "erasers"); items.add("paper"); for (String x : items) {…

The correct answer is D. A runtime exception is thrown. Option D is correct because items.add(1, "pen") is called on an empty list. The index-based overload add(int index, E element) requires the index to satisfy 0 <= index <= size; since the list has size 0 at that point, index 1 is out of bounds and an IndexOutOfBoundsException is…

Arrays and Logic

Question

Given the code fragment:
List<String> items = new ArrayList<> ();
items.add(1, "pen");
items.add(2, "pencil");
items.add(3, "erasers");
items.add("paper");
for (String x : items) {
System.out.print(x + " ");
}
What is the result?

Options

Apen pencil erasers paper
Bpaper pen pencil erasers
CA compilation error occurs.
DA runtime exception is thrown.

How the community answered

(34 responses)

A
3% (1)
B
3% (1)
C
9% (3)
D
85% (29)

Explanation

Option D is correct because items.add(1, "pen") is called on an empty list. The index-based overload add(int index, E element) requires the index to satisfy 0 <= index <= size; since the list has size 0 at that point, index 1 is out of bounds and an IndexOutOfBoundsException is thrown immediately, before any items are printed.

Options A and B are wrong because execution never reaches the loop - the exception halts the program on the second line. Option C is wrong because the code compiles cleanly; both add(int index, E element) and add(E element) are valid overloads of List, and the integer literals are unambiguous.

Memory tip: When you see list.add(index, value) on a fresh ArrayList, ask yourself "is that index within [0, size]?" An empty list only accepts index 0 - any higher index is a runtime trap, not a compile-time one.

Topics

#ArrayList#IndexOutOfBoundsException#Method overloading#Collection API

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