string and final confusion

[Akshay Rawal]

String s1 = "str";
String s2 = "ing";

String concat = s1 + s2 ; // i think it returns new object and it is not placed in string constant pool..just guessing dont know exact reason..

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



final String s1 = "str";
final String s2 = "ing";

String concat = s1 + s2 ; // it reurns new object and place in string constant pool..just a guess..

System.out.println(concat == "string"); //true.....can anyone please help me the reason why both are giving unexpected answer

[Dieter Quickfend]

This depends on the compiler and is at this point an unreliable comparison. If the compiler gets both references from the String pool it will return true, if it doesn't, it will return false. To compare their true value, you must use equals method. Using equals operator will return true for some compilers (in reality, probably all compilers), but it's not an obligation according to the specification and should thus never be relied upon.

[Henry Wong]

Akshay Rawal wrote:String s1 = "str";
String s2 = "ing";

String concat = s1 + s2 ; // i think it returns new object and it is not placed in string constant pool..just guessing dont know exact reason..

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



final String s1 = "str";
final String s2 = "ing";

String concat = s1 + s2 ; // it reurns new object and place in string constant pool..just a guess..

System.out.println(concat == "string"); //true.....can anyone please help me the reason why both are giving unexpected answer

First read this... https://coderanch.com/t/454384/java/java/compile-time-constant

The two points of that topic that are important are... one, the concatenation of two compile time constants is a compile time constant. So, when you concat two compile time constant variables, it creates a compile time constant, which in turn, if it is used, the compiler will place it into the string pool.

And two, there are requirements that determine what is a compile time constant variable. In the first example, those requirements are not met. In the second example, those requirements are met.

Henry

[Henry Wong]

Akshay Rawal wrote:
final String s1 = "str";
final String s2 = "ing";

String concat = s1 + s2 ; // it reurns new object and place in string constant pool..just a guess..

System.out.println(concat == "string"); //true.....can anyone please help me the reason why both are giving unexpected answer

Interestingly, no. Or perhaps ... almost.

s1, s2, "str", "ing", and "string" are all compile time constants. The compiler knows what every one of these values are at compile time. And hence, it is like the code you had were ...

String concat = "string";

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

So, yes, you can can argue that an object is created and placed in the string pool... but the object is not created from the concatenation operation.

Also note that the "str" and "ing" objects are *not* created at all !!

Henry

[Akshay Rawal]

ok..now i got...
1>"str" and "ing" are compile time constant literal....
2>String s1,s2 are not compile time constant variable..
3>and final string s1,s2 are compile time constant variable

[Joanne Neal]

Henry Wong wrote:Also note that the "str" and "ing" objects are *not* created at all !!

Are you sure about that ? The variables s1 and s2 still exist, so surely they are pointing at Strings "str" and "ing" respectively.

[Henry Wong]

Joanne Neal wrote:

Henry Wong wrote:Also note that the "str" and "ing" objects are *not* created at all !!

Are you sure about that ? The variables s1 and s2 still exist, so surely they are pointing at Strings "str" and "ing" respectively.

Correct. The variables exist -- but interestingly, the fields don't exist. And if the variables were directly used, the objects would exist too.

Henry