Question

Modify the given Java code to handle non-integer inputs gracefully. Instead of termination, display the message "Input is not valid" and prompt the user to re-enter a number.

publicstaticvoidmain(String[] args) {
intcountValidNumber=0;
int num;
for(inti=0;i<100;i++)
{
System (""Enter a Positive Number:"");
String s1
=JOptionPane.showInputDialoge("enter number",null);
num = Integer.parseInt(s1);
if(num>0){countValidNumber++;
}
}
System.out.println(""Valid Numbers Are:""+countValidNumber);
}

Answer 1

The given Java code needs to be modified to handle non-integer inputs gracefully by displaying a 'Input is not valid' message and prompting the user to re-enter a number. This can be done using a try-catch block to catch the NumberFormatException.

Step-by-step explanation:

The given Java code needs to be modified to handle non-integer inputs gracefully. Instead of terminating when a non-integer input is provided, we need to display the message 'Input is not valid' and prompt the user to re-enter a number.

One way to handle this is by using a try-catch block. We can wrap the line "num = Integer.parseInt(s1);" inside a try block and catch the NumberFormatException that is thrown when a non-integer input is provided.

Here is the modified code:

public static void main(String[] args) {
int countValidNumber = 0;
int num;
for (int i = 0; i < 100; i++) {
System.out.println("Enter a Positive Number:");
String s1 = JOptionPane.showInputDialog("enter number", null);
try {
num = Integer.parseInt(s1);
if (num > 0) {
countValidNumber++;
}
} catch (NumberFormatException e) {
System.out.println("Input is not valid");
}
}
System.out.println("Valid Numbers Are: " + countValidNumber);
}