Does java create a new Object when a primitive type is passed to a method that expects a Wrapper of a primitive type?

Question

Say that we have:

public method(Integer s) {
   ....
}

Java admits the invocation of that method doing:

method(7);

or

int i = 7;
method(i);

Does the JVM create a new Integer when doing that? If I invoke the method several times with the same int, am I creating several Integers, or a "cached" version of the wrapper object is used?

Thank you.

Answer 1

Does the JVM create a new Integer when doing that?

Not for the example you've given, no - because it's in the range [-128, 127]. If your example had been 157 instead, the answer would have been implementation-specific.

From section 5.1.7 of the JLS:

If the value p being boxed is true, false, a byte, or a char in the range \u0000 to \u007f, or an int or short number between -128 and 127 (inclusive), then let r₁ and r₂ be the results of any two boxing conversions of p. It is always the case that r₁ == r₂.

Ideally, boxing a given primitive value p, would always yield an identical reference. In practice, this may not be feasible using existing implementation techniques. The rules above are a pragmatic compromise. The final clause above requires that certain common values always be boxed into indistinguishable objects. The implementation may cache these, lazily or eagerly. For other values, this formulation disallows any assumptions about the identity of the boxed values on the programmer's part. This would allow (but not require) sharing of some or all of these references.

This ensures that in most common cases, the behavior will be the desired one, without imposing an undue performance penalty, especially on small devices. Less memory-limited implementations might, for example, cache all char and short values, as well as int and long values in the range of -32K to +32K.