🔥 Why 2 == 2 is true but 2000 == 2000 is false in Java? 🤯 Integer a = 2; Integer b = 2; System.out.println(a == b); // true ✅ Integer x = 2000; Integer y = 2000; System.out.println(x == y); // false ❌ Looks weird, right? How can 2000 == 2000 be false? 🤔 Here’s what’s happening 👇 🔹 Java has an Integer Cache for values in the range -128 to 127. 👉 So when you write Integer a = 2; Integer b = 2;, both refer to the same cached object → true. 🔹 But numbers outside this range (like 2000, 500, or -200) are not cached by default. 👉 Each statement creates a new object, meaning different references → false. 💡 Fun Fact: You can extend the cache range (upper bound only) using a JVM option: -Djava.lang.Integer.IntegerCache.high=1000 But unless configured, the default range remains -128 to 127. ⚡ Key Takeaway: ✅ Use == for primitives ✅ Use .equals() for wrapper classes These subtle details can trip even experienced developers — and yes, they often come up in interviews too 😅 Have you ever been surprised by this Java quirk? Drop your thoughts below 👇 #Java #CodingTips #JavaInterview #InterviewPreparation #ProgrammingConcepts #CoreJava #SoftwareEngineering #LearningEveryday #CodeWisdom #DevelopersLife
On objects, .equals() should be used == operaror might have weird behavior
Integer Cache due to its range -2^7 to 2^7
x.equals(y);
Nice question
Its due to Integer caching😁
its beyond the limit because java maintain integer cache between -128 to 127 so thats 2000=2000 false