do ... while

Figure 163. A do ... while loop Slide presentation 
int sum = 0, value;
do {
  System.out.print(
    "Enter value, 0 to terminate: ");
  value = scan.nextInt();
  sum += value;
} while (0 != value);
System.out.println("Sum: " + sum);
Enter value, 0 to terminate: 3
Enter value, 0 to terminate: 1
Enter value, 0 to terminate: 0
Sum: 4
A do ... while loop


Figure 164. do ... while syntax Slide presentation 
do
  (block | statement)
while (booleanExpression);


exercise No. 70

Even or odd?

Q:

Write an application which asks for integer values telling its user whether a given value is even or odd. Providing the special value 0 shall terminate your application: 

Enter an integer (0 to terminate): 77
    77 is an odd number
Enter an integer (0 to terminate): -3
    -3 is an odd number
Enter an integer (0 to terminate): 26
    26 is an even number
Enter an integer (0 to terminate): 0
    0 is an even number
Goodbye!

Tip

  1. Use the modulo “%” operator.

  2. Choose an appropriate loop type with respect to your application's termination on user input «0».

A:

We obviously need a loop to ask for further input unless the last entered value was 0. In any case the loop's statement will be executed at least once: 

Enter an integer (0 to terminate): 0
    0 is an even number
Goodbye!

A do ... while(...) rather than a while(...) loop is thus appropriate. It allows for entering the loop's body before any check is about to happen: 

final Scanner scanner = new Scanner(System.in));

  int userInput;
    do {
      System.out.print("Enter an integer (0 to terminate): ");
      userInput = scanner.nextInt();
      if (0 == userInput % 2) {
        System.out.println("    " + userInput + " is an even number");
      } else {
        System.out.println("    " + userInput + " is an odd number");
      }
    } while (0 != userInput);
  System.out.println("Goodbye!");
}

exercise No. 71

Square root approximation

Q:

Derived from the Newton–Raphson method we can approximate a given value a 's square root a by the following recursively defined series:

Start:
x 0 = a 2
Recursion step:
x n + 1 = 1 2 ⁢ ( x n + a x n )

Create a program which interactively asks a user for a positive value and calculate it's square root like e.g.: 

Enter a non-negative value: 2.0
The square root of 2.0 is close to 1.414213562373095
It's square is 1.9999999999999996

Tip

Due to the limited precision of machine arithmetics you may continue until your approximation value no longer changes.

A:

We introduce two variables x_current and x_next referring to x n and x n + 1 respectively: 

final Scanner scan = new Scanner(System.in));

System.out.print("Enter a non-negative value: ");
final double a = scan.nextDouble();

double x_next = a / 2, x_current;

do {
  x_current = x_next;                          // Save current approximation value
  x_next = (x_current + a / x_current) / 2;    // Calculate next series value
} while (x_next != x_current);                 // Did we get any closer?

System.out.println("The square root of " + a + " is close to " + x_next);
System.out.println("It's square is " + x_next * x_next);
Prev  Up  Next
while  Home  for