Reputation: 362
This is the program that checks if the input integer is binary or not, now I need to create a loop that will prompt the user to renter integers until binary number is entered.
import java.util.Scanner;
public class BinaryNumbers {
public static void main(String[] args) {
int value, userValue;
int binaryDigit = 0, notBinaryDigit = 0;
Scanner scan = new Scanner(System.in);
System.out.println("Please enter positive integers: ");
userValue = scan.nextInt();
value = userValue;
while (userValue > 0) {
if ((userValue % 10 == 0) || (userValue % 10 == 1)) {
binaryDigit++;
} else {
notBinaryDigit++;
}
userValue = userValue / 10;
}
if (notBinaryDigit == 0) {
System.out.println(value + " is a Binary Number.");
} else {
System.out.println(value + " is not a Binary Number.");
}
}
}
Upvotes: 1
Views: 567
Reputation: 4377
Why not using Regular Expressions?
All those checking on user inputs by assuming them as numbers (by calling Scanner#nextInt
or Scanner#nextFloat
or ...) are very breakable. How can be so sure user won't enter anything wrong?
It's better to hold the user input in a String
variable and check it against being binary integer (which has to be 32 bit at most as defined in many languages such as java) using Regex is more safe:
import java.util.Scanner;
public class CheckBinaryInteger {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
boolean isValid = false;
String userInput = "";
do {
System.out.print("Please Enter a binary integer: ");
userInput = sc.next();
isValid = userInput != null && !userInput.trim().isEmpty()
&& userInput.matches("[01]{1,32}");//assume every digit as one bit
if(!isValid)
System.out.println("invalid binary integer entered! ");
}while(!isValid);
System.out.println("Valid input: "+userInput);
}
}
Hope this helps.
Upvotes: 1
Reputation: 448
import java.util.Scanner;
public class BinaryNumbers {
public static void main(String[] args) {
int value, userValue;
int binaryDigit = 0, notBinaryDigit = 0;
Scanner scan = new Scanner(System.in);
while(true)
{
System.out.println("Please enter positive integers: ");
userValue = scan.nextInt();
value = userValue;
while (userValue > 0)
{
if ((userValue % 10 == 0) || (userValue % 10 == 1))
binaryDigit++;
else
notBinaryDigit++;
userValue = userValue / 10;
}
if (notBinaryDigit == 0)
{
System.out.println(value + " is a Binary Number.");
break;
}
else
System.out.println(value + " is not a Binary Number.");
}
}
}
Upvotes: 0
Reputation: 310
import java.util.Scanner;
class calculatePremium
{
public static void main(String[] args) {
int value, userValue;
int binaryDigit = 0, notBinaryDigit = 0;
Scanner scan = new Scanner(System.in);
while(true) /*use a while loop to iterate till you get the binary number*/
{
binaryDigit = 0; notBinaryDigit = 0;
System.out.println("Please enter positive integers: ");
userValue = scan.nextInt();
value = userValue;
while (userValue > 0) {
if ((userValue % 10 == 0) || (userValue % 10 == 1)) {
binaryDigit++;
} else {
notBinaryDigit++;
}
userValue = userValue / 10;
}
if (notBinaryDigit == 0) {
System.out.println(value + " is a Binary Number.");
break; /* breaks out of loop when gets the correct input */
} else {
System.out.println(value + " is not a Binary Number.\n");
}
}
}
}
You just need to use a loop till you get a binary number. Hope it helps.
Upvotes: 0
Reputation: 2543
import java.util.Scanner;
public class BinaryNumbers {
public static void main(String[] args) {
int value, userValue;
Scanner scan = new Scanner(System.in);
while(true){
int binaryDigit = 0, notBinaryDigit = 0;
System.out.println("Please enter positive integers: ");
userValue = scan.nextInt();
value = userValue;
while (userValue > 0) {
if ((userValue % 10 == 0) || (userValue % 10 == 1)) {
binaryDigit++;
} else {
notBinaryDigit++;
}
userValue = userValue / 10;
}
if (notBinaryDigit == 0) {
System.out.println(value + " is a Binary Number.");
return; //does the trick ;)
} else {
System.out.println(value + " is not a Binary Number.");
}
}
}
}
A simple return can end the program then and there :)
Upvotes: 1