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

It is important to realize that the first if will make 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 will be picked. The variable suffix will get "nd", and the false branch will be completely skipped.

QUESTION 9:

The expression (count+1) will always be 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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