Common Array Algorithms

22. Array Equality

Answer:
if ( array.length > 0 )
{
  . . .
  System.out.println("The average is: " + total / array.length  );
}      
else
  System.out.println("The array contains no elements." );

Array Equality

Are the following two arrays equal?

int[] arrayA = { 1, 2, 3, 4 };
int[] arrayB = { 7, 8, 9};

Obviously not.

Are these two arrays equal?

int[] arrayC = { 1, 2, 3, 4 };
int[] arrayD = { 4, 3, 2, 1 };

Less obvious, but ordinarily they would not be regarded as equal.

What about these two arrays: are they equal?

int[] arrayE = { 1, 2, 3, 4 };
int[] arrayF = { 1, 2, 3, 4 };

Here, it depends on what you mean by "equal". The object referred to by the variable arrayE is not the same object that is referred to by the variable arrayE. The "alias detector" == returns false.

                     
class ArrayEquality
{
  public static void main ( String[] args )
  {
    int[] arrayE = { 1, 2, 3, 4 };
    int[] arrayF = { 1, 2, 3, 4 };
    
    if (arrayE==arrayF)
      System.out.println( "Equal" );
    else
      System.out.println( "NOT Equal" );      
  }
}

Output:
NOT Equal

You have seen this situation before, with two separate String objects containing the same characters:

class StringEquality
{
  public static void main ( String[] args )
  {
    String stringE = new String( "Red Delicious");
    String stringF = new String( "Red Delicious");
    
    if (arrayE==arrayF)
      System.out.println( "Equal" );
    else
      System.out.println( "NOT Equal" );      
  }
}

Output:
NOT Equal

There are two individual objects, so the object reference in stringE is not == to the object reference in stringF.

Question 22:
Confusion Alert! (review of Chapter 43)
    String stringG = "Red Delicious" ;
    String stringH = "Red Delicious" ;

    if (arrayG==arrayH)
      System.out.println( "One Literal" );
    else
      System.out.println( "NOT Equal" );         
What does the above print?