EXP32-J. Do not use the equal and not equal operators to compare boxed primitives

The values of boxed primitives cannot be compared using the == and != operators by default. This is because these are interpreted as reference comparison operators.

Autoboxing can automatically wrap the primitive type to the corresponding wrapper object. Some care should be taken during this process, especially when performing comparisons. The Java Language Specification [[JLS 05]] explains this point clearly:

If the value p being boxed is true, false, a byte, a char in the range \u0000 to \u007f, or an int or short number between -128 and 127, then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.

Noncompliant Code Example

This noncompliant example (adopted from [[Bloch 09]]), defines a Comparator with a compare() method. The compare() method accepts two boxed primitives as arguments. Note that primitive integers are also accepted by this declaration as they are appropriately autoboxed. The main issue is that the == operator is being used to compare the two boxed primitives. This however, compares their references and not the actual values.

static Comparator<Integer> cmp = new Comparator<Integer>() {
  public int compare(Integer i, Integer j) {
    return i < j ? -1 : (i == j ? 0 : 1);
  } 
};

Compliant Solution

To be compliant, use any of the four comparison operators <, >, <= and >=. The == and != operators should not be used to compare boxed primitives.

public int compare(Integer i, Integer j) {
  return i < j ? -1 : (i > j ? 1 : 0) ;
}

Noncompliant Code Example

This code uses == to compare two integer objects. According to EXP03-J. Do not compare String objects using equality or relational operators, for == to return true for two object references, they must point to the same underlying object. Results of using the == operator in this case will be misleading.

public class TestWrapper2 {
 public static void main(String[] args) {
 
  Integer i1 = 100;
  Integer i2 = 100;
  Integer i3 = 1000;
  Integer i4 = 1000;
  System.out.println(i1==i2);
  System.out.println(i1!=i2);
  System.out.println(i3==i4);
  System.out.println(i3!=i4);
 
 }
}

These comparisons generate the output sequence: true, false, false and true. The cache in the Integer class can only make the integers from -127 to 128 refer to the same object, which explains the output of the above code. To avoid making such mistakes, use equals instead of == to compare wrapper classes (See EXP03-J for further details).

Compliant Solution

Using object1.equals(object2) only compares their values. Now, the results will be true, as expected.

public class TestWrapper2 {
 public static void main(String[] args) {
  
  Integer i1 = 100;
  Integer i2 = 100;
  Integer i3 = 1000;
  Integer i4 = 1000;
  System.out.println(i1.equals(i2));
  System.out.println(i3.equals(i4));

 }
}

Noncompliant Code Example

Sometimes a list of integers is desired. Recall that the type parameter inside the angle brackets of a list cannot be of a primitive type. It is not possible to form an ArrayList<int>. With the help of the wrapper classs and autoboxing, storing primitive integer values in ArrayList<Integer> becomes possible.

In this noncompliant code example, it is desired to count the integers of arrays list1 and list2. As class Integer can only cache integers from -127 to 128, when an int number is beyond this range, it is autoboxed into the corresponding wrapper type. The == operator returns false when these distinct wrapper objects are compared. As a result, the output of this example is 0.

import java.util.ArrayList;
public class Wrapper {
 public static void main(String[] args) {
   // Create an array list of integers, where each element 
   // is greater than 127
   ArrayList<Integer> list1 = new ArrayList<Integer>();
   for(int i=0;i<10;i++)
     list1.add(i+1000);
   // Create another array list of integers, where each element
   // is the same as the first list
   ArrayList<Integer> list2 = new ArrayList<Integer>();
   for(int i=0;i<10;i++)
     list2.add(i+1000);
 
   int counter = 0;
   for(int i=0;i<10;i++)
     if(list1.get(i) == list2.get(i)) 
       counter++;
   // print the counter
   System.out.println(counter);
 }
}

If it were possible to expand the cache inside Integer (cache all the integer values -32K-32K, which means that all the int values may be autoboxed to the same Integer object), then the results may have differed.

Compliant Solution

This compliant solution uses the equals() method for performing comparisons of wrapped objects. It produces the correct output 10.

public class TestWrapper1 {
 public static void main(String[] args) {
   // Create an array list of integers, where each element
   // is greater than 127
   ArrayList<Integer> list1 = new ArrayList<Integer>();
   for(int i=0;i<10;i++)
     list1.add(i+1000);
   // Create another array list of integers, where each element
   // is the same as the first one
   ArrayList<Integer> list2 = new ArrayList<Integer>();
   for(int i=0;i<10;i++)
     list2.add(i+1000);
 
   int counter = 0;
   for(int i=0;i<10;i++)
     if(list1.get(i).equals(list2.get(i))) counter++;
 
   System.out.println(counter);
 }
}

Risk Assessment

Using the equal and not equal operators to compare boxed primitives can lead to erroneous comparisons.

Automated Detection

TODO

Related Vulnerabilities

Search for vulnerabilities resulting from the violation of this rule on the CERT website.

References

[[Bloch 09]] 4. "Searching for the One"
[[Pugh 09]] Using == to compare objects rather than .equals

EXP31-J. Avoid side effects in assertions      03. Expressions (EXP)      04. Scope (SCP)