Java String pool object creation

Question

I have doubts that whether my concepts are clear in stringpool. Please study the following set of code and check if my answers are correct in number of objects created after the following set of statements:-

1)

String s1 = "abc";
String s2 = "def";
s2 + "xyz";

2)

 String s1 = "abc";
 String s2 = "def";
 s2 = s2 + "xyz";

3)

String s1 = "abc";
String s2 = "def";
String s3 = s2 + "xyz";

4)

String s1 = "abc";
String s2 = "def";
s2 + "xyz";
String s3 = "defxyz";

As per what i know conceptually, in all the 4 cases above, there will be 4 objects created after the execution of each set of the lines.

Tell us why, for example, in number 3 why there are 4 objects. — djna
– djna, Commented Aug 22, 2011 at 7:06
@djna: True. The compiler is free to use only three objects, since s2 + "xyz" can be evaluated at compile-time. — Thilo
– Thilo, Commented Aug 22, 2011 at 7:07
@Thilo, In the last one there are 4 distinct literals. I suppose we are not considering dead code elimination. :) — Ray Toal
– Ray Toal, Commented Aug 22, 2011 at 7:12

Peter Lawrey · Accepted Answer · 2011-08-22 08:01:04Z

8

You cannot have an expression like s2 + "xyz" on its own. Only constants are evaluated by the compiler and only string constants are automatically added to the String literal pool.

e.g.

final String s1 = "abc"; // string literal
String s2 = "abc"; // same string literal

String s3 = s1 + "xyz"; // constants evaluated by the compiler 
                        // and turned into "abcxyz"

String s4 = s2 + "xyz"; // s2 is not a constant and this will
                        // be evaluated at runtime. not in the literal pool.
assert s1 == s2;
assert s3 != s4; // different strings.

answered Aug 22, 2011 at 8:01

Peter Lawrey

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Comments

user207421 · Accepted Answer · 2011-08-23 01:30:28Z

1

Why do you care? Some of this depends on how aggressively the compiler optimizes so there is no actual correct answer.

answered Aug 23, 2011 at 1:30

user207421

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