Is there any bug in java.lang.String literal implementation

Question

Consider the following program:
class StringTest
     public static void main(String[] args)
          long freeMemInitially=Runtime.getRuntime().freeMemory();
          String s="Amarnath";
          //char data[] = {'A', 'm', 'a','r','n','a','t','h'};
     //String s = new String(data);
          long usedMemAfterFirstString=freeMemInitially-Runtime.getRuntime().freeMemory();
          System.out.println("usedMemAfterFirstString:= "+usedMemAfterFirstString);
Run the program it will return 0. But why? As the java docs and java specification says that string litterals is the reference to the instance of java.lang.String object.Then where does this string have been saved except the java runtime heap?
Message was edited by:
Amarnath_Bhattacharya

Answer 1

basically, the simple explanation is this:
the string literal "usedMemAfterFirstString" and the string literal "Amarnath" are both instantiated BEFORE the main method is called.
hence, no object creation is done between the first and second invocations of Runtime.getRuntime().freeMemory().
If you take a closer look at the spec, it will confirm this.
- Adam