Answer:
import java.util.*;
public class num2 {
public static void main(String[] args) {
//Set the Value of the argument
int val = 5;
//Call the method
int returnedNum = testInteger(val);
//Use a While to continously call the method as long as the returned value is not 0
while(returnedNum != 0){
int rt = testInteger(val);
//Break out of loop when the user enters correct value and the return is 0
if(rt == 0){
break;
}
}
}
//The Method definition
public static int testInteger(int val){
//Prompt and receive user input
System.out.println("Enter a number");
Scanner in = new Scanner(System.in);
int userNum = in.nextInt();
//Check different values using if and else statements
if(userNum > val){
System.out.println("Too High");
return 1;
}
else if(userNum < val){
System.out.println("Too Low");
return -1;
}
else{
System.out.println("Got it");
return 0;
}
}
}
Explanation:
This is solved with Java Programing Language
Pay close attention to the comments provided in the code
The logic is using a while loop to countinually prompt the user to enter a new number as long as the number entered by the user is not equal to the value of the argument already set to 5.
Answer:
def high_low(val):
number = int(input("Enter a number: "))
if number > val:
print("Too high")
return 1
elif number < val:
print("Too low")
return -1
elif number == val:
print("Got it!")
return 0
val = 7
while True:
if high_low(val) == 0:
break
Explanation:
The code is in Python.
Create a function called high_low that takes one parameter, var
Ask the user for a number. Check the number. If the number is greater than 1, smaller than 1 or equal to 1. Depending on the number, print the required output and return the value.
Set the value to a number. Create a while loop iterates until the user enters the correct number, when the return value of the function is equal to 0, that means the user entered the correct number. Otherwise, it keeps asking the number and prints the appropriate message.
