Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
262 views
in Technique[技术] by (71.8m points)

java - Boxed Primitives and Equivalence

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?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

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


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...