Answer:

The first (index 0) element in an array will never be larger than the maximum element in that array. (It might be the maximum element of the array; but this is fine.)

max = array[0];

This code assumes that there is at least one element in the array. If there is not, the java system will detect that the index "0" is out of bounds and throw an exception.

Classic Bug (1): Initializing Max

Classic Bug (1): It is tempting to initialize max to something you think is much less than the maximum, like -999. However, in doing so you make a risky assumption about the data. Perhaps the array holds the national debt for each of the the last ten years. All these values are negative, and -999 would easily exceed the true maximum.

Here is the program so far. It has initialized max to the first element in the array. The for loop is set up to look at every element in the array, starting with the first element in cell 0, to see if that element is larger than the current maximum.

class MaxAlgorithm
{

  public static void main ( String[] args ) 
  {

    int[] array =  { -20, 19, 1, 5, -1, 27, 19, 5 } ;
    int   max;

    // initialize the current maximum
 
    max = array[0];

    // scan the array
    for ( int index=0; index < array.length; index++ )
    {
     
      if (  )  // examine the current element

        max = array[ index ];        // if it is the largest so far, change max

    }
      
    System.out.println("The maximum of this array is: " + max );

  }
}

It would be OK (and perhaps preferable) to initialize index at 1 since index 0 has already been used to initialize max. But the program as written works fine (as soon as you fill in the blank.)

QUESTION 9:

Complete the program.