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Is it safe to use !=/== with Character?

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java

Is it safe to use ==/!= while comparing Character?

Character being a boxed type is it safe to use ==/!= while comparing Character types?

  public static void main(String[] args) {

        Character c1 = 'd';
        Character c2 = (char) getInt();

        System.out.println(c1 == c2);
    }

    public static int getInt() {

        return 100;
    }

The following works as expected (true). However, are there cases where comparing Character with same value using == would lead to false? (Hence, do we have to use '.equals()' while comparing boxed primitive types?

like image 646
user7858768 Avatar asked Oct 08 '21 17:10

user7858768


2 Answers

No, it's not safe. You must use equals().

Demonstration:

System.out.println(Character.valueOf('Ü') == Character.valueOf('Ü'));
// -> false

Note that if you use autoboxing or Character.valueOf(), then some characters (ASCII characters) are cached and the same Character instance is reused, so == may return true for the same value:

System.out.println(Character.valueOf('A') == Character.valueOf('A'));
// -> true (on my machine)

But it doesn't work for all characters, and it won't work if you call the deprecated new Character(...) explicitly.

like image 108
Alex Shesterov Avatar answered Nov 15 '22 17:11

Alex Shesterov


tl;dr

Use code points, not char/Character.

"d".codePointAt( 0 ) == 100  // true.

Details

The Answer by Alex Shesterov is correct. But bigger picture, you should not be using Character objects.

Character is broken

The Character class is a wrapper class for the primitive type char. The char/Character type is legacy as of Java 2, and is essentially broken. As a 16-bit value, it is physically incapable of representing most characters.

For example, try running:

 System.out.println( Character.valueOf( '😷' ) ) ;

Code points

Instead, when working with individual characters, use code point integer numbers. In Java that means using the int/Integer type.

If you look around classes such as String, StringBuilder, and Character you will find codePoint methods.

Let's revise your code snippet. We will change the names to be more descriptive. We switch out Character and char usage for mere int primitive integers. As such, we can compare our int values using == or !=.

package work.basil.text;

public class App7
{
    public static void main ( String[] args )
    {
        int codePointOf_LATIN_SMALL_LETTER_D = "d".codePointAt( 0 ); // Annoying zero-based index counting, not ordinal.
        int codePoint2 = getInt();

        boolean sameCharacter = ( codePointOf_LATIN_SMALL_LETTER_D == codePoint2 );  // Comparing `int` primitives with double-equals. 
        System.out.println( sameCharacter );
    }

    public static int getInt ()
    {
        return 100;  // Code point 100 is LATIN SMALL LETTER D, `d`. 
    }
}

When run:

true

Of course, if you use auto-boxing or otherwise mix the wrapper class Integer with the primitive int, then the same explanation in that other Answer applies here too.

  • Compare Integer object to Integer object by calling Integer#equals or Objects.equals( Object a , Object b ) rather than using ==.
  • Compare int primitive to int primitive using == or calling Integer.compare( int x , int y ).
like image 27
Basil Bourque Avatar answered Nov 15 '22 18:11

Basil Bourque