Two string reference are referring to same string but when comparing, result return false.

Question

The value of s2 and s3 is "ab" and this string is stored in string pool. As per doc s2==s3 should return true but it is returning false. Why?

 public class Sample {
      public static void main(String[] args) {
      String s="a";
      String s1="b";
      String s2=s+s1;
      String s3="ab";
      System.out.println(s2==s3);
   }
}

Why do you think `String s2=s+s1;` stores the result in the string pool? It does not. — Eran, Jun 28 '18 at 05:53
Bit of advice: forget about the String pool. There is nothing to be gained by correct understanding of its workings and everything to lose by incorrect understanding. — Kevin Anderson, Jun 28 '18 at 06:20
@Downvoters Why the downvotes? A perfectly sound and non-trivial question. — lexicore, Jun 28 '18 at 06:57

Sweeper · Accepted Answer · 2018-06-28T06:54:17.087

According to the Java Language Specification §15.18.1

15.18.1 String Concatenation Operator +

If only one operand expression is of type String, then string conversion (§5.1.11) is performed on the other operand to produce a string at run time.

The result of string concatenation is a reference to a String object that is the concatenation of the two operand strings. The characters of the left-hand operand precede the characters of the right-hand operand in the newly created string.

The String object is newly created (§12.5) unless the expression is a constant expression (§15.28).

+ will always produce a new string, unless the expression is constant. One example of a constant expression is "a" + "b". Your expression is not constant because it contains non-final variables.

This is stated even more clearly in §12.5:

Execution of a string concatenation operator + (§15.18.1) that is not part of a constant expression (§15.28) always creates a new String object to represent the result.

Technically, if the variables were constants, the expression would still be a constant expression. — kumesana, Jun 28 '18 at 06:49
@kumesana Variables are variables, constants are constants. But I get you. Corrected. — Sweeper, Jun 28 '18 at 06:50
Not according to Java. As far as Java is concerned, a constant is, among other things, a variable. — kumesana, Jun 28 '18 at 06:51

score -1 · Answer 2 · answered Jun 28 '18 at 06:47

Sure both s2 and s3 equal to "ab".

And sure the "ab" String is into the String pool, because you had that String as a literal in your code. Which also means that s3 is taken from the String pool since you're assigning the said literal to it.

But why would you think that s + s1 would pull the resulting String out of the String pool? It wouldn't.

Two string reference are referring to same string but when comparing, result return false.

2 Answers2