Clarification on Strings

[Phani Burra]

Hello Ranchers,

I have quick question wrt Strings in Java.

 String x="a"+"b";
String y="a"+"b";
if(x==y)
System.out.println("x=y");
String y1="a";
String y2=y1+"b";
if(x==y2)
System.out.println("x=y2");   

In the above code the first condition x==y is satisfied but the second condition x==y2 give output false but both x and y2 have "ab" in them. My understanding of strings is that these string objects are stored in the string constant pool memory and whenever a string is created it is first checked in the pool memory and when it is found then the object is referenced to the string in the pool. But this doesn't justify the above condition. So please can anyone explain what happens in the memory for the above situation. Thanks in advance.

[Irina Goble]

Only x, y and y1 will be in the String constant pool. For y2 Java will create a new Stirng object in normal memory, because "y1 +"b"" is not a compile-time constant expression. The String constant pool is for String literals only. SeeJLS:

15.18.1 String Concatenation Operator +
If only one operand expression is of type String, then string conversion is performed on the other operand to produce a string at run time. The result is a reference to a String object (newly created, unless the expression is a compile-time constant expression (�15.28))that is the concatenation of the two operand strings.

[ January 31, 2008: Message edited by: Irina Goble ]
[ January 31, 2008: Message edited by: Irina Goble ]

[Phani Burra]

Hey thank you very much.