String vs Object

[jacob deiter]

String k="f";

is same as

String k=new String("f"); ??


If that is the case ,if a class is declared as follow

Class A
{

String z="z";
String y="y";
String x="x";


}

It says that Class A contain three String Object?

Class A
{

Object z="z";
Object y="y";
Object x="x";


}


which one is best if I do not want any String operation like concatation,substr,...etc

[Seetharaman Venkatasamy]

jacob deiter wrote:
String k="f";
is same as
String k=new String("f"); ??

No.

jacob deiter wrote:
If that is the case ,if a class is declared as follow
Class A
{
String z="z";
String y="y";
String x="x";
}

It says that Class A contain three String Object?

Class A
{

Object z="z";
Object y="y";
Object x="x";


}

which one is best if I do not want any String operation like concatation,substr,...etc

if you dont want string operation,then why you create it?

[jacob deiter]

No

please explain this

[Nitish Bangera]

String s = "x" creates string object at compile time and the object is referenced from the string pool.
String s = new String("x") create string object at runtime.

Well Object o = "x" is allowed in terms of polymorphism. You can use such types if you want to use one type only and then later can downcast when you really want to use that specific methods.

[Seetharaman Venkatasamy]

Nitish Bangera wrote:String s = "x" creates string object at compile time

Object Creation always in runtime

[Seetharaman Venkatasamy]

jacob deiter wrote:

No

please explain this

example,

when you say

    
String s = "abc";
String t = "abc";

there is only one object "abc" created in heap and two reference for the objects are placed in the string constant pool.

but here

    
String s = new String("abc");
String t = new String("abc");

there are 2 objects "abc" and "abc" are created in the heap though there are meaningfully same.

and s and t are pointing to the different object .

so always good practice is use String s = "abc" definition

Hope this helps

[Nitish Bangera]

String literals are compile time constants.

See here

[Vijitha Kumara]

Nitish Bangera wrote:String s = "x" creates string object at compile time

Well, actually all objects are created at runtime for the String literals it's when the class gets loaded by the JVM.

[Vijitha Kumara]

seetharaman venkatasamy wrote:.. two reference for the objects are placed in the string constant pool.

No. SCP contains references to unique String literals. So whenever your program make a reference to that String object it is replaced with the reference to the object referenced from SCP. Check StringLiterals.

[Balu Sadhasivam]

Nitish Bangera wrote:String literals are compile time constants.

See here

Yes literals are compile constants , that doesn't mean String objects are created at compile time.

Literals are just "abc" or 'c' 1 or 133.9 ...

[shivendra tripathi]

doesn't mean String objects are created at compile time

If you are creating string as literal then certainly string object will be created at compile time. Please refer this http://www.javaranch.com/journal/200409/Journal200409.jsp#a1 for further clarification.

[Paul Sturrock]

shivendra tripathi wrote:

doesn't mean String objects are created at compile time

If you are creating string as literal then certainly string object will be created at compile time. Please refer this http://www.javaranch.com/journal/200409/Journal200409.jsp#a1 for further clarification.

Can you point out the bit in this article that says instances of objects are created a compile time? How are any instances of an object created at a point in the procedings where there is no JVM involved, and hence no heap to store these on?

[shivendra tripathi]

Sorry for that. string object will be created at class loading time for string literal.

[jacob deiter]

seetharaman venkatasamy wrote:

jacob deiter wrote:

No

please explain this

example,

when you say

    
String s = "abc";
String t = "abc";

there is only one object "abc" created in heap and two reference for the objects are placed in the string constant pool.

but here

    
String s = new String("abc");
String t = new String("abc");

there are 2 objects "abc" and "abc" are created in the heap though there are meaningfully same.

and s and t are pointing to the different object .

so always good practice is use String s = "abc" definition

Hope this helps

but the below code

  
		String a="abc";
		String b="abc";
		String c=new String ("abc");
System.out.println(a.hashCode());
System.out.println(b.hashCode());
System.out.println(c.hashCode());  

return the same hash code value???

[Campbell Ritchie]

jacob deiter wrote:but the below code
. . .
return the same hash code value???

Read Object#equals(), Object#hashCode(), String#equals() and String#hashCode(), which should explain why you get the same hash code.

[Nitish Bangera]

Strings and Wrappers have overridden equals and hashcode methods. This is the reason when we use Strings and Wrappers in maps, we don't need to do implement any methods.

[shaiksha vali]

What is mean by String Mutable and Immutable

[Vijitha Kumara]

shaiksha vali wrote:What is mean by String Mutable and Immutable

I don't get that statement . Strings are immutable
Now this is a bit off topic. Mutable means that you can change the data contained in the object once created. For String once you create an object of String you cannot change the value it holds (i.e: Immutable).

[Max Rahder]

To get back to the original question:

You can reference methods of an object based on the variable type. For example, if you store a String in a String variable, you can run String methods on it. If you store the String in a variable of type Object, then you can only run Object methods.

For example:

String s = "hi there";
System.out.println(s.charAt(1)); // This compiles because charAt(int) is a String method

Object o = "hi there";
System.out.println(o.charAt(1)); // This does NOT compile because charAt(int) is NOT an Object method
System.out.println(o.toString()); // This compiles, because toString() is an Object method