How is String concatenation working in following

Question

As per the following How Java String pool works when String concatenation? conversation,

String a = "hello world!"; String b = "hello" + " world!"; boolean compare = (a == b);

The compare should be true which is correct. However, I've the following code

String s1 = "This is"; String s2 = " a new String"; String s3 = s1 + s2; String s4 = "This is a new String";

On comparing System.out.printf("s3 == s4:%s\n", s3 == s4);//always false It is always false. My understanding is that on concatenation of s1 & s2, s3 will create a string in pool and when s4 is created it will point to the same pool location. But this is not the case. I've tried this with different JDKs including 7, 8 and 14 and the results are consistent.

"s3 will create a string in pool" no. Runtime-evaluated strings are only added to the pool if you explicitly request it, with intern(). — Andy Turner
– Andy Turner, Commented May 4, 2020 at 9:28

Community · Accepted Answer · 2020-06-20 09:12:55Z

This is how the behaviour regarding pooling strings is specified in the Java Language Specification (emphasis mine):

3.10.5 String Literals

[...]

Moreover, a string literal always refers to the same instance of class String. This is because string literals - or, more generally, strings that are the values of constant expressions (§15.28) - are "interned" so as to share unique instances, using the method String.intern.

Only constant expressions of type String are pooled. "This is a new String", and "hello" + " world!" are constant expressions. The compiler can evaluate these expressions at compile time. s1 + s2 is not a constant expression.

So when executing s1 + s2, a new string is created. But note that another string, which just so happens to have the same characters as the new string, is in the string pool, because you used a string literal to initialise s4.

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