What will be the output of this code? var i = 1234; if (i == "1234") { System.out.println("Alpha"); } else if (i == 1234) { System.out.println("Beta"); } else { System.out.println("Gamma"); }

Question

Options

• ADoes not compile
• BPrints "Alpha" only
• CPrints "Alpha" followed by "Beta"
• DPrints "Beta" only
• EPrints "Gamma"
• FThrows an Exception

• A
  84% (21)
• B
  4% (1)
• C
  8% (2)
• E
  4% (1)

Explanation

Option A is correct because Java is a strictly-typed language - comparing an int (which is what var i = 1234 infers) to a String literal using == is a compile-time type error; the compiler rejects incompatible types before the program ever runs. Options B, C, D, and E are all wrong because they assume the program reaches runtime, which it never does. Options B and C further assume Java performs implicit type coercion like JavaScript (turning 1234 into "1234" for comparison), but Java never coerces primitives to String silently. Option E would only be reachable if both prior conditions were false, which is moot since compilation fails first. Option F (throws an exception) is wrong because type mismatches in Java surface at compile time, not as runtime exceptions.

Memory tip: Java = "type mismatch = won't compile." When you see == comparing two operands of unrelated types (like int vs String), stop - the compiler stops there too. If you'd need a cast or .equals() to make the comparison valid, the raw == won't pass the compiler.

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