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As a general rule, use the Object.equals() method to check whether two objects are abstractly equal to each other. Reserve use of the equality operators == and != for testing whether two references specifically refer to the same object (this is reference equality). See also guideline MET13-J. Classes that define an equals() method must also define a hashCode() method.

When operating on This also applies to numeric boxed types (for example,Byte, Character, Short, Integer, Long, Float, and Double), although the numeric relational operators (such as <, <=, >, and >=) produce results that match those provided for arguments of the equivalent primitive numeric types. Specifically, the JLS requires auto-unboxing in this case, which results in comparison of the numeric values contained in the boxed objects. ( See JLS Section 5.6.2, "Binary Numeric Promotion".) But when both arguments of an equality operator (for example, == or !=) are of a numeric boxed type, the operation is a reference comparison rather than the anticipated numeric comparison, which can produce unexpected results. (See guideline EXP03-J. Avoid the equal and not equal operators when comparing values of boxed primitives for more information.)

Noncompliant Code Example

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public class ExampleComparison {
  public static void main(String[] args) {
    String onestr1 = new String("one");
    String twostr2 = new String("one");
    boolean result;
    // test below is redundant in this case, but required for full generality
    if (onestr1 == null) { 
      result = twostr2 == null;
    }
    else {
      result = onestr1 == twostr2;
    }
    System.out.println(result); // prints false
  }
}

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public class GoodComparison {
  public static void main(String[] args) {
    String onestr1 = new String("one");
    String twostr2 = new String("one");
    boolean result;
    if (onestr1 == null) {
      result = two(str2 == null);
    } else {
      result = onestr1.equals(twostr2);
    }
    System.out.println(result);
  }
}

Compliant Solution (String.intern())

Reference equality produces the desired result when comparing string literals (for example, String one = "one"; and String two = "two";) or when the intern method has been used on both strings.behaves like abstract object equality when comparing two strings that are each the result of the String.intern() method. When a task requires keeping only one copy of each string in memory, as well as performing quick repeated string comparisons, this compliant solution may be used.

public class GoodComparison {
  public static void main(String[] args) {
    String onestr1 = new String("one");
    String twostr2 = new String("one");
    boolean result;
    if (onestr1 != null) {
      onestr1 = onestr1.intern();
    }
    if (twostr2 != null) {
      twostr2 = twostr2.intern();
    }
    result = onestr1 == twostr2;

   System.out.println(result);
  }
}

Use this approach with care; performance and clarity could be better served by use of code that applies the Object.equals() approach and lacks a any dependence on reference equality.

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EXP01-EX2: Use reference equality to determine whether two references point to the same object instance.

Risk Assessment

Using reference equality to compare objects may lead to unexpected results.

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