Create Java String Using ” ” or Constructor?


In Java, a string can be created by using two methods:

String x = "abc";
String y = new String("abc");

What is the difference between using double quotes and using constructor?

1. Double Quotes vs. Constructor

This question can be answered by using two simple code examples.

Example 1:

String a = "abcd";
String b = "abcd";
System.out.println(a == b);  // True
System.out.println(a.equals(b)); // True

a==b is true because a and b are referring to the same string literal in the method area. The memory references are the same.

When the same string literal is created more than once, only one copy of each distinct string value is stored. This is called "string interning". All compile-time constant strings in Java are automatically interned.

Example 2:

String c = new String("abcd");
String d = new String("abcd");
System.out.println(c == d);  // False
System.out.println(c.equals(d)); // True

c==d is false because c and d refer to two different objects in the heap. Different objects always have different memory references.

This diagram illustrate the two situations above:
constructor vs double quotes Java String - New Page

2. Run-Time String Interning

Thanks to LukasEder (his comment below):

String interning can still be done at run-time, even if two strings are constructed with constructors:

String c = new String("abcd").intern();
String d = new String("abcd").intern();
System.out.println(c == d);  // Now true
System.out.println(c.equals(d)); // True

3. When to Use Which

Because the literal "abcd" is already of type String, using constructor will create an extra unnecessary object. Therefore, double quotes should be used if you just need to create a String.

If you do need to create a new object in the heap, constructor should be used. Here is a use case.

Category >> Basics >> Diagram >> Java  
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  • Walt Corey

    This is how one writes code nobody can understand. Good way to be top of the list come next layoff. I knew a guy did the same crap wit C++, so ‘highly optimized” nobody could read it, and it also didn’t work but broke in mysterious ways. Rather than seeing how obtuse you can code, see how readable you can. Readability is FAR more cost effective than so-called highly optimized code.

  • Ritter Liu

    Thank you for your share :P

  • Charly

    You are worng, In JDK1.6 and JDK1.7 return true. Test rourself

  • 智 陶

    String c = new String(“abcd”).intern();
    String d = new String(“abcd”).intern();
    System.out.println(c == d); // True in JDK1.7 while false in JDK 1.6.
    correct me if i am wrong

  • Feng Sun

    good to know, thanks

  • LukasEder

    When the same string literal is created more than once, JVM store only one copy of each distinct string value. This is called “string interning“.

    That’s not entirely correct. In your example, the compiler will already make sure that the same string constant is effectively referenced. String interning can still be done at runtime, even if two strings are constructed with constructors:

    String c = new String(“abcd”).intern();
    String d = new String(“abcd”).intern();
    System.out.println(c == d); // Now true
    System.out.println(c.equals(d)); // True