📌 A Small Java Trick Many Developers Don’t Know Most developers know that Java is pass-by-value. But here’s something interesting : What will be the output of this code? public class Test { static void modify(String str) { str = "World"; } public static void main(String[] args) { String s = "Hello"; modify(s); System.out.println(s); } } Take a second and think about it 👀 🧠 Many People Expect: World Because the method tries to modify the value. But the actual output is: Hello 🔎 Why Does This Happen? Java always passes arguments by value. When modify(s) is called: - A copy of the reference is passed - The method changes only the local copy - The original reference still points to "Hello" So the original variable remains unchanged. 🎯 Key Takeaway Java is strictly pass-by-value, even for objects. What gets passed is the value of the reference, not the object itself. This small detail explains many confusing bugs during debugging. Follow for more Java insights, interview questions, and system design concepts. #Java #Programming #SoftwareEngineering #JavaDeveloper #CodingInterview
Classic Java gotcha! Combine this strict pass-by-value behavior with String immutability, and you get the foundation for thread-safe operations in highly concurrent environments. Mastering these fundamentals is exactly what saves us from nasty bugs when scaling complex distributed systems.
Omg is this on linkedin? This is literally the first class of learning java... Literally level zero. Should be in a media focused on people that really are having first programming lesson not on linkedin where people are in majority working professionals Seriously dude gotta understand the audience