String problem.

[dmjpro]

Plz guys look at this code carefully.

[code=java]
String a = "abc";
String b = "abc";
String c = new String("abc");
System.out.prin tln(a == b);
System.out.prin tln(a == c);

//and the output is true and false.

[/code]
why not the output is false and false?
plz explain.

kind regards.
dmjpro.

[r035198x]

Plz guys look at this code carefully.

[code=java]
String a = "abc";
String b = "abc";
String c = new String("abc");
System.out.prin tln(a == b);
System.out.prin tln(a == c);

//and the output is true and false.

[/code]
why not the output is false and false?
plz explain.

kind regards.
dmjpro.

Think references and remember the String pool.
In
[CODE=java] String b = "abc";[/CODE] no new oject is created because the literal "abc" already exists in the String pool so the reference b is merely made to point to the same memory that a is pointing to. In

[CODE=java]String c = new String("abc");[/CODE] The new keyword forces a new object to be created in memory disregarding what is in the String pool.

[sumittyagi]

Plz guys look at this code carefully.

[code=java]
String a = "abc";
String b = "abc";
String c = new String("abc");
System.out.prin tln(a == b);
System.out.prin tln(a == c);

//and the output is true and false.

[/code]
why not the output is false and false?
plz explain.

kind regards.
dmjpro.

This is due to string pool.
whenever you define a string, that string is dropped in a pool of strings. When you define another string with same char sequence, then new string object is not created, the same object kept in the pool is returned.

But if you create a string object by using new keyword, then java have no option other than creating new object as you directed it to do so.

That's why first two references are same but third one is not.

[dmjpro]

I came to know a good thing .

thanks for ur suggestions.

kind regards.
dmjpro.

[JosAH]

Also have a look at the String.intern() method; this method adds a String to
the String pool.

kind regards,

Jos

[dmjpro]

That means all time the String Object is not pooled every time.
right???

so the first print will not be true every time.
right???

if not then why String.intern() ?????
plz help.

kind regards.
dmjpro.

[JosAH]

That means all time the String Object is not pooled every time.
right???

so the first print will not be true every time.
right???

if not then why String.intern() ?????
plz help.

kind regards.
dmjpro.

String literals such as "abc" are always pooled and the following will always be true:
[code=java]
String a= "abc";
String b= "abc"
System.out.prin tln(a == b); // true
[/code]

The following will always be false:
[code=java]
String c= new String("abc");
System.out.prin tln(a == c || b == c); // false
[/code]
And the following will be true again:
[code=java]
c= c.intern();
System.out.prin tln(a == c && b == c); // true
[/code]

kind regards,

Jos

[dmjpro]

Now the picture comes more clearer.
Thanks for ur suggestions.

kind regards.
dmjpro.