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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T09:27:14+00:00 2026-05-13T09:27:14+00:00

So I was asked this question today. Integer a = 3; Integer b =

  • 0

So I was asked this question today.

Integer a = 3;
Integer b = 2;
Integer c = 5;
Integer d = a + b;
System.out.println(c == d);

What will this program print out? It returns true. I answered it will always print out false because of how I understood auto (and auto un) boxing. I was under the impression that assigning Integer a = 3 will create a new Integer(3) so that an == will evaluate the reference rather then the primitive value.

Can anyone explain this?

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  1. Editorial Team
    Editorial Team
    2026-05-13T09:27:14+00:00Added an answer on May 13, 2026 at 9:27 am

    Boxed values between -128 to 127 are cached. Boxing uses Integer.valueOf method, which uses the cache. Values outside the range are not cached and always created as a new instance. Since your values fall into the cached range, values are equal using == operator.

    Quote from Java language specification:

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

    http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.7

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