Java String pool and type casting

Question

My question is in regard to the way Java handles String literals. It\'s quite clear from the Java Language Specs (JLS) that String literals are being implicitly interned - in ot

Answer 1

Seems like because you reference an object here final String stringObject = (String) object;, this is no longer a 'compile-time' constant, but a 'run-time' constant. The first example from here eludes to it with the part:

String s = "lo";
String str7 = "Hel"+ s;  
String str8 = "He" + "llo"; 
System.out.println("str7 is computed at runtime.");     
System.out.println("str8 is created by using string constant expression.");    
System.out.println("    str7 == str8 is " + (str7 == str8));  
System.out.println("    str7.equals(str8) is " + str7.equals(str8));

The string str7 is computed at runtime, because it references another string that is not a literal, so by that logic I assume despite that face that you make stringObject final, it still references an object, so cannot be computed at compile time.

And from the java lang spec here, it states:

"The string concatenation operator + (§15.18.1) implicitly creates a new String object when the result is not a compile-time constant expression (§15.28). "

I cannot find any examples where a cast can be used, except, for this terrible, terrible example:

System.out.println(hello == "hel" + ( String ) "lo");

Which hardly has any logical use, but maybe the part about a string cast was included because of the above case.

Answer 2

Casting to Object is not allowed in a compile time constant expression. The only casts permitted are to String and primitives. JLS (Java SE 7 edition) section 15.28:

> - Casts to primitive types and casts to type String

(There's actually a second reason. object isn't final so cannot possibly by considered a constant variable. "A variable of primitive type or type String, that is final and initialized with a compile-time constant expression (§15.28), is called a constant variable." -- section 4.12.4.)