Answer:

If the digits cannot be converted into a 32-bit int an Exception is thrown. This limits the input to the range of an int, roughly plus or minus 2 billion.

`hasNextInt()`

import java.util.Scanner;
import java.io.*;

public class ManySquares
{
  public static void main (String[] args) throws IOException
  { 
    File    file = new File("myData.txt");   // create a File object
    Scanner scan = new Scanner( file );      // connect a Scanner to the file
    int num, square;      

    while( scan.hasNextInt() )   // is there more data to process? 
    {
      num = scan.nextInt();
      square = num * num ;      
      System.out.println("The square of " + num + " is " + square);
    }
    scan.close();
  }
}

Most files contain a great deal of data. Practical programs must process data from a file with a loop of some sort. Here is a program that reads a text file that contains many integers and writes the square of each one. The loop reads integers from the file one by one until the end of file is reached or until input unsuitable for nextInt() is encountered.

However, the program computes a correct square only for input values less than the square root of 2 billion, roughly 46000. Integers larger than this result in overflow. (A better program would check for this.)

The hasNextInt() method returns true if the next set of characters in the input stream can be read in as an int. If they can't be read as an int, or they are too big for an int or if the end of the file has been reached, then it returns false. hasNextInt() itself does not consume any characters.

Answer:

hasNextInt()

QUESTION 7:

`hasNextInt()`