How Does Declaring Strings as `final` Affect `==` Comparison in Java?

Understanding the Impact of Final on String Comparison in Java

In Java, the equality of strings is typically determined using the equals() method. However, the == operator can also be used for comparison, but its behavior differs when dealing with final strings.

When a String variable is declared final and initialized with a compile-time constant expression, it becomes a constant variable and its value is inlined by the compiler. This inlining leads to a significant difference in the comparison using ==.

Consider the following code example:

String str1="str";
String str2="ing";
String concat=str1+str2;

System.out.println(concat=="string"); // false

The comparison in this example returns false because concat is a new String object created at runtime by concatenating str1 and str2.

Now, let's modify the code by declaring str1 and str2 as final:

final String str1="str";
final String str2="ing";
String concat=str1+str2;

System.out.println(concat=="string"); // true

In this case, the comparison returns true. This is because when str1 and str2 are declared final, they become constant variables and are inlined at compile time. The concatenation of the two strings is also computed at compile time, resulting in a constant String object with the value "string."

The difference arises because string literals are interned in Java. Interning involves storing a unique instance of a String object in a pool for future reference. When a final String is inlined at compile time and compared with a string literal, the references to the String objects in the intern pool are directly compared, resulting in true.

In summary, when dealing with final strings declared with compile-time constant expressions, the comparison using == will return true if the concatenated value is equal to the other string literal, due to the interning of String objects at compile time.

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