String Pool(String and StringBuffer)

Question

On some blog and site even on stack overflow, I can see similar ans for below question

    String s = new String("Test");

Will create two object as literal "Test" will take place in pool.

So how StringBuffer perform in below case.

    StringBuffer sb = new StringBuffer("BufferTest");

Is the literal "BufferTest" also take place in pool ?

if yes then how StringBuffer save String garbage collection ?

Have a look at this as well :- stackoverflow.com/questions/2909848/… — Goyal Vicky
– Goyal Vicky, Commented Aug 15, 2015 at 15:28

user3453226 · Accepted Answer · 2015-08-15 14:58:51Z

0

"BufferTest" will take place in the pool because any String object does that.

answered Aug 15, 2015 at 14:58

user3453226

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Stephen C · Accepted Answer · 2015-08-15 16:52:13Z

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Is the literal "BufferTest" also take place in pool ?

Yes.

All literal strings are help in the string pool/

if yes then how StringBuffer save String garbage collection ?

It doesn't need to. A String corresponding to a string literal remains reachable so long as the class (or classes) that defined it is in existence / loaded.

answered Aug 15, 2015 at 16:52

Stephen C

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2 Comments

Amit Over a year ago

But As per below code I think with every append call StringBuffer loosing reference of first. <br/> ' StringBuffer sb = new StringBuffer();' ' sb.append("Test");' ' System.out.println(sb.toString());' sb.append("Test1"); System.out.println(sb.toString());

Stephen C Over a year ago

Like I said, the StringBuffer doesn't need to keep a reference to it. The reference for "Test" is placed in the string pool when the code is loaded, not when it is executed.

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