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    String s1 = "abc";
    String s2 = "def";
    String s3 = "abcdef";

    String s4 = s1 + s2;

    String s5 = "abc" + "def";

    System.out.println(s3==s4); //gives false
    System.out.println(s3==s5); //gives true

Can anyone explain why s3==s4 is giving false and s3==s is giving true? unable to understand the reason and internal working of what is actually happening here?

When I do concatenation of string like in s5, the logic that "abcdef" value is already there in string pool because of s3, for s5==s3 we will get true as s5 will also be pointed to the same s3 object present in stringpool in the heap.

Going the same logic, i.e. value already there is stringpool so new ref. variable will also point to same object, I thought s3==s4 would give true as s3 is already there is string pool, so when s4 is created it will point to same "abcdef" i.e. the s3 value in string pool.....but s3==s4 is giving false.....can someone please explain what am I missing here if s3==s5 is working and giving true...why isn't s3==s4 working and giving true as well?

Vishal Singh
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  • This comes from an inappropriate way of comparing. – deHaar Mar 01 '23 at 08:33
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    `"abc" + "def"` is semantically identical to `"abcdef"` at compile time; `s1 + s2` is calculated at runtime, because at least one of `s1` and `s2` isn't a compile-time constant expression. Try making both `s1` and `s2` `final`. – Andy Turner Mar 01 '23 at 08:35
  • @AndyTurner........yes when I make both s1 and s2 as final I get true.....can you explain more what this final is actually doing? – Vishal Singh Mar 01 '23 at 08:40
  • `final` makes `s1` and `s2` compile-time constant expressions (because they are final *and* initialized with compile-time constant values). Therefore `s1 + s2` is also a compile-time constant expression. – Andy Turner Mar 01 '23 at 08:43

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