🚀 Day 4 — The Java Trick That Even Seniors Miss! 😮 Every Java dev thinks they understand method overloading... But this one small detail breaks their confidence 👇 public class OverloadTest { void show(Object o) { System.out.println("Object"); } void show(String s) { System.out.println("String"); } public static void main(String[] args) { OverloadTest test = new OverloadTest(); test.show(null); } } 🧠 Question: What will be the output of this code? Will it print: 1️⃣ Object 2️⃣ String 3️⃣ Compile-time error Most developers get this wrong! 😅 💬 Drop your answer in the comments 👇 Let's see how many of you actually know how Java resolves overloaded methods at compile time! #Java #CodingChallenges #SpringBoot #JavaDevelopers #InterviewQuestion #Day4Challenges #cbfr
When multiple overloaded methods match, it selects the most specific type. output will be : String
String
The compiler chooses the one with the most specific parameter type. So the output will be String.