[ Exercises | Chapter Index | Main Index ]

## Solution for Programmming Exercise 2.6

This page contains a sample solution to one of the exercises from Introduction to Programming Using Java.

### Exercise 2.6:

Suppose that a file named "testdata.txt" contains the following information: The first line of the file is the name of a student. Each of the next three lines contains an integer. The integers are the student's scores on three exams. Write a program that will read the information in the file and display (on standard output) a message the contains the name of the student and the student's average grade on the three exams. The average is obtained by adding up the individual exam grades and then dividing by the number of exams.

Discussion

TextIO can be used to read data from a file; this is discussed in Subsection 2.4.5. To read data from a file named "testdata.txt", all you need to do is say TextIO.readFile("testdata.txt"). From then on, the input functions in TextIO will read from the file instead of reading data typed in by the user. (Note that this assumes that the file is in the same directory with the program.) In this case, we can use TextIO.getln() to read the student's name from the first line of the file, and then we can read the exam grades by calling TextIO.getln() three times.

The average should be computed as a value of type double. Don't forget that if you divide an integer by an integer in Java, the result is an integer and the remainder of the division is discarded. To get the correct average in this case, the program divides the sum of the three grades by 3.0 rather than by 3.

One final technicality is that simply outputting a double value might print out something like 83.333333333333333. By default, all significant digits in the number are output. In this case, one digit after the decimal point is probably sufficient. The program uses formatted output to achieve this. The format string "The average grade for %s was %1.1f" is used to format the name and the average. The name is substituted for the format specifier %s, which means that the name is printed as a string, with no extra spaces. The average is substituted for %1.1f, which means that the average is printed as a floating point number with no extra spaces and with 1 digit after the decimal point.

You might want to run this program with no data file, or with a data file that is not in the correct format, to see what happens. (The program will crash and print an error message.)

The Solution

    public class FindAverage {

       public static void main(String[] args) {

           String name;     // The student's name, from the first line of the file.
           int exam1, exam2, exam3;     // The student's grades on the three exams.
           double average;  // The average of the three exam grades.

           TextIO.readFile("testdata.txt");  // Read from the file.

           name = TextIO.getln();  // Reads the entire first line of the file.
           exam1 = TextIO.getlnInt();
           exam2 = TextIO.getlnInt();
           exam3 = TextIO.getlnInt();

           average = ( exam1 + exam2 + exam3 ) / 3.0;

           System.out.printf("The average grade for %s was %1.1f", name, average);
           System.out.println();

       }

    }

[ Exercises | Chapter Index | Main Index ]