Integer i2 = new Integer(100);
Byte b2 = new Byte((byte)100);
Also check equals method logic for Byte here..
Hiram Nascimento wrote:c = b; // Okay. b is "boxed" and its reference is assigned to c. Right?
// Does c get "unboxed" or b "boxed"?
print(c == b); // true
Try the following to see which occurs:
// Why false?
print(c.equals(120)); // false
Because 120 is auto boxed into an Integer, and a Byte and an Integer are never equals.
// Is c "unboxed" or the integer literal "boxed"?
print(c == 120); // true
See my above little test code.
When I do "if(c == b)", the == is doing a comparison of the references of these vars. The Byte.equals() method returns true if the equals parameter is a "instanceof" Byte. That's how it was implemented.
The autoboxing is related the Wrapper classes of primitive types. Each primitive type in Java (int, byte, long, float, etc) has a corresponding (wrapper) class (Integer, Byte, Long, Float, etc) to be used when an object is expected instead of a primitive value/type.
Before Java 5.0, the process to handle the interaction of primitive types and its wrappers was "manual"... Autoboxing came to make things easier.
you should read more about Wrapper classes and Autoboxing to understand better that process. Not too complex. But some aspects are implementation details you need to know.
Not easy to summarize all this in one post, but read more about Wrapper classes and Autoboxing and tell us your questions. I'm still studying that too.
cchetan jain wrote:hi guys..sorry frnds..i m not cleared with the concept of boxing and unboxing concept yet..what is the concept of equals and ==.??please explain this..thanks in advance
Use equals if both operands are Objects (wrapper instances).
Use == if at least one is a primitive.
Use the valueOf method combined with equals if one is a primitive and the other is a wrapper instance that may perhaps be null to prevent NullPointerExceptions: