Given: import java.util.function.Supplier; public class MyLambda { public static void main(String[] args) { int i = 25; Supplier<Integer> foo = () -> i; i++; System.out.println(foo.get()); } } Which…

Question

Options

• AThe code compiles but does not print any result.
• BThe code prints 25.
• CThe code does not compile.
• DThe code throws an exception at runtime.

• A
  6% (1)
• B
  13% (2)
• C
  75% (12)
• D
  6% (1)

Explanation

Option C is correct because Java requires variables captured by a lambda to be effectively final - meaning their value cannot change after being assigned. The line i++ modifies i after it is captured by the lambda, which violates this rule, causing a compile-time error ("local variable i defined in an enclosing scope must be effectively final").

Why the distractors are wrong:

• A is wrong because the code never reaches runtime - the compiler rejects it outright.
• B is wrong for the same reason; there is no execution, so nothing prints.
• D is wrong because the failure happens at compile time, not at runtime.

Memory tip: Think of lambda captures as a "snapshot" - Java needs to guarantee the value won't change out from under the lambda, so any variable a lambda touches must be frozen (final or effectively final) from the moment of capture onward. If you see a post-capture mutation (i++, i = newValue, etc.), the code won't compile.

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