Question

The following code is supposed to compute the floor of the square root of its input value x. (Note: The floor of a number n is the largest integer less than or equal to n.) // Computes and writes floor(sqrt(x)) for // an input value x >= 0. public static void main(String args[]) { int x; // input value Scanner input = new Scanner(System.in); // initialize int result = 0; // will equal floor of sqrt(x) int temp1 = 1; int temp2 = 1; // read input x = input.nextInt(); // compute floor while (temp1 < x) { ++result; temp2 += 2; temp1 += temp2; } // end while System.out.println("The floor of the square root of "+ x +" is " + result); } // end main This program contains an error. a. What output does the program produce when x = 64? b. Run the program and remove the error. Describe the steps that you took to find the error. c. How can you make the program more user friendly and fail-safe?

          The following code is supposed to compute the floor of the square root of its input value x.
(Note: The floor of a number n is the largest integer less than or equal to n.)
// Computes and writes floor(sqrt(x)) for
// an input value x >= 0.
public static void main(String args[]) {
int x; // input value
Scanner input = new Scanner(System.in);
// initialize
int result = 0; // will equal floor of sqrt(x)
int temp1 = 1;
int temp2 = 1;
// read input
x = input.nextInt();
// compute floor
while (temp1 < x) {
++result;
temp2 += 2;
temp1 += temp2;
} // end while
System.out.println("The floor of the square root of "+ x +" is " + result);
} // end main
This program contains an error.
a. What output does the program produce when x = 64?
b. Run the program and remove the error. Describe the steps that you took to find the error.
c. How can you make the program more user friendly and fail-safe?

The following code is supposed to compute the floor of the square root of its input value x.
(Note: The floor of a number n is the largest integer less than or equal to n.)
// Computes and writes floor(sqrt(x)) for
// an input value x >= 0.
public static void main(String args[])
int x; // input value
Scanner input = new Scanner(System.in);
// initialize
int result = 0; // will equal floor of sqrt(x)
int temp1 = 1;
int temp2 = 1;
// read input
x = input.nextInt();
// compute floor
while (temp1 < x)
++result;
temp2 += 2;
temp1 += temp2;
// end while
System.out.println("The floor of the square root of "+ x +" is " + result);
// end main
This program contains an error.
a. What output does the program produce when x = 64?
b. Run the program and remove the error. Describe the steps that you took to find the error.
c. How can you make the program more user friendly and fail-safe?

Added by Cameron V.

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