Fill in the two blanks so that the nested if's make the correct choice.

A good answer might be:

if ( count+1  == 2  )
  suffix = "nd";
else
  if ( count+1  == 3  )
    suffix = "rd";
  else
    suffix = "th";

System.out.println( "Enter the " + 
    (count+1) + suffix + " integer (enter 0 to quit):" );


How the Three-way Choice Works

The first if makes a choice between its true branch and its false branch. Its false branch is complicated, but that doesn't matter. Each branch of an if statement can be as complicated as needed. Here is the fragment again, with the first if's true branch in blue and its false branch in red: 

if ( count+1  == 2  )
  suffix = "nd";
else
  if ( count+1 == 3  )
    suffix = "rd";
  else
    suffix = "th";

System.out.println( "Enter the " + (count+1) + suffix + " integer (enter 0 to quit):" );

For example, if (count+1) is equal to 2, the true branch is picked. The variable suffix gets "nd", and the false branch is completely skipped.

QUESTION 9:

The expression (count+1) is always greater than 1. For each of the following possible values of (count+1) which branch of the FIRST if will be executed?

    value of count+1which branch? 
if ( count+1  == 2  )
  suffix = "nd";
else
  if ( count+1 == 3  )
    suffix = "rd";
  else
    suffix = "th";
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