Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Answer:
import java.util.Scanner;
import java.lang.*;
class Main {
public static void main(String[] args) {
boolean exit = false;
Scanner reader = new Scanner(System.in);
while(!exit) {
char c = '?';
while( !Character.isLetter(c)) {
System.out.print("Enter a letter or * to exit: ");
c = reader.next().charAt(0);
if (c == '*') {
exit = true;
break;
}
if (!Character.isLetter(c)) {
System.out.printf("Error: %c is not a letter!\n", c);
}
}
if (Character.isLowerCase(c)) {
System.out.printf("%c in uppercase is %c\n", c, Character.toUpperCase(c));
} else if (Character.isUpperCase(c)) {
System.out.printf("%c in lowercase is %c\n", c, Character.toLowerCase(c));
}
}
reader.close();
}
}
Explanation:
I decided to use the asterisk (*) as exit character.
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.