I've read this answer about how to check if a string is interned in Java, but I don't understand the following results:
String x = args[0]; // args[0] = "abc";
String a = "a";
String y = a + "bc";
System.out.println(y.intern() == y); // true
But if I declare a string literal:
String x = "abc";
String a = "a";
String y = a + "bc";
System.out.println(y.intern() == y); // false
Besides, without any string literal, the args[0] seems to be directly interned:
// String x = "abc";
String y = args[0];
System.out.println(y.intern() == y); // true (???)
// false if the first line is uncommented
Why does y.intern() == y change depending on whether x is a literal or not, even for the example when the command-line argument is used?
I know literal strings are interned at compile time, but I don't get why it affects in the previous examples. I have also read several questions about string interning, like String Pool behavior, Questions about Java's String pool and Java String pool - When does the pool change?. However, none of them gives a possible explanation to this behaviour.
Edit:
I wrongly wrote that in third example the result doesn't change if String x = "abc"; is declared, but it does.
It is because y.intern() gives back y if the string was not interned before. If the string already existed, the call will give back the already existing instance which is most likely different from y.
However, all this is highly implementation dependent so may be different on different versions of the JVM and the compiler.