Difference between string object and string litera

Question

This question already has an answer here:

• What is the difference between “text” and new String(“text”)? 9 answers

What is the difference between

String str = new String("abc");

and

String str = "abc";

Answer 1

When you use a string literal the string can be interned, but when you use new String("...") you get a new string object.

In this example both string literals refer the same object:

String a = "abc"; 
String b = "abc";
System.out.println(a == b);  // true

Here, 2 different objects are created and they have different references:

String c = new String("abc");
String d = new String("abc");
System.out.println(c == d);  // false

In general, you should use the string literal notation when possible. It is easier to read and it gives the compiler a chance to optimize your code.

Answer 2

As Strings are immutable, when you do:

String a = "xyz"

while creating the string, the JVM searches in the pool of strings if there already exists a string value "xyz", if so 'a' will simply be a reference of that string and no new String object is created.

But if you say:

String a = new String("xyz")

you force JVM to create a new String reference, even if "xyz" is in its pool.

For more information read this.

Answer 3

Some disassembly is always interesting...

$ cat Test.java 
public class Test {
    public static void main(String... args) {
        String abc = "abc";
        String def = new String("def");
    }
}

$ javap -c -v Test
Compiled from "Test.java"
public class Test extends java.lang.Object
  SourceFile: "Test.java"
  minor version: 0
  major version: 50
  Constant pool:
const #1 = Method  #7.#16;  //  java/lang/Object."<init>":()V
const #2 = String  #17;     //  abc
const #3 = class   #18;     //  java/lang/String
const #4 = String  #19;     //  def
const #5 = Method  #3.#20;  //  java/lang/String."<init>":(Ljava/lang/String;)V
const #6 = class   #21;     //  Test
const #7 = class   #22;     //  java/lang/Object
const #8 = Asciz   <init>;
...

{
public Test(); ...    

public static void main(java.lang.String[]);
  Code:
   Stack=3, Locals=3, Args_size=1
    0:    ldc #2;           // Load string constant "abc"
    2:    astore_1          // Store top of stack onto local variable 1
    3:    new #3;           // class java/lang/String
    6:    dup               // duplicate top of stack
    7:    ldc #4;           // Load string constant "def"
    9:    invokespecial #5; // Invoke constructor
   12:    astore_2          // Store top of stack onto local variable 2
   13:    return
}

Answer 4

According to String class documentation they are equivalent.

Documentation for String(String original) also says that: Unless an explicit copy of original is needed, use of this constructor is unnecessary since Strings are immutable.

Look for other responses, because it seems that Java documentation is misleading :(

Answer 5

String is a class in Java different from other programming languages. So as for every class the object declaration and initialization is

String st1 = new String();

or

String st2 = new String("Hello"); 
String st3 = new String("Hello");

Here, st1, st2 and st3 are different objects.

That is:

st1 == st2 // false
st1 == st3 // false
st2 == st3 // false

Because st1, st2, st3 are referencing 3 different objects, and == checks for the equality in memory location, hence the result.

But:

st1.equals(st2) // false
st2.equals(st3) // true

Here .equals() method checks for the content, and the content of st1 = "", st2 = "hello" and st3 = "hello". Hence the result.

And in the case of the String declaration

String st = "hello";

Here, intern() method of String class is called, and checks if "hello" is in intern pool, and if not, it is added to intern pool, and if "hello" exist in intern pool, then st will point to the memory of the existing "hello".

So in case of:

String st3 = "hello";
String st4 = "hello"; 

Here:

st3 == st4 // true

Because st3 and st4 pointing to same memory address.

Also:

st3.equals(st4);  // true as usual

Answer 6

String s = new String("FFFF") creates 2 objects: "FFFF" string and String object, which point to "FFFF" string, so it is like pointer to pointer (reference to reference, I am not keen with terminology).

It is said you should never use new String("FFFF")