Unable to handle infinite loop when using Scanner for input

Question

I have this piece of code :

import java.util.*;
class HelloWorld {
    public static void main(String args[]) {
        Scanner ob = new Scanner(System.in);
        int t = ob.nextInt(); //no. of Test Cases
                ob.next(); // for next input whether exception occurs or not
        int a = 0, c = 0;
        for (int j = 0; j < t; j++)
        {
            a = 0; c = 0;
            String str = ob.nextLine();
            String [] spstr = str.split("\\s+");
            try
            {
                for (int i=0 ; i<spstr.length ; i++)
                {
                    if(spstr[i].equals("")) {i--;}
                    else {
                        c = c + Integer.parseInt(spstr[i]);
                        }
                }
                System.out.println(c);
            } catch (Exception e) {
                System.out.println("Invalid Input");
                }
        }
    }
}

What this code do is add any no.s of integers in a single line. Before doing this, I have a Test Case int t. This decides how many inputs must be taken. But this results to an infinite loop even when I am entering integer value.

I have seen this post: How to handle infinite loop caused by invalid input using Scanner which have many answers on how to get rid of this. I have followed the answers, but I have not yet solved this issue.

Note: When I use int t=5;, it works fine. But in this case too, if exception is caught twice, same thing happens.

Please tell me how to solve this infinite loop error ?

Thanks in Advance :)

The answer says to add a line `ob.next();` but doing that also doesn't helped me — Somendra Meena, Jun 03 '16 at 19:30

score 1 · Answer 1 · answered Jun 03 '16 at 20:33

To begin with, proper indentation helps make code easier to read.

class HelloWorld {
    public static void main(String args[]) {
        Scanner ob = new Scanner(System.in);
        int t = ob.nextInt(); //no. of Test Cases
        ob.next(); // for next input whether exception occurs or not
        int a = 0, c = 0;
        for (int j = 0; j < t; j++) {
            a = 0;
            c = 0;
            String str = ob.nextLine();
            String[] spstr = str.split("\\s+");
            try {
                for (int i = 0; i < spstr.length; i++) {
                    if (spstr[i].equals("")) {
                        i--;
                    } else {
                        c = c + Integer.parseInt(spstr[i]);
                    }
                }
                System.out.println(c);
            } catch (Exception e) {
                System.out.println("Invalid Input");
            }
        }
    }
}

There's several problems.

int t = ob.nextInt(); //no. of Test Cases
ob.next(); // for next input whether exception occurs or not

I don't know what you're hoping to accomplish here. That is not what the linked answer has as a solution. The linked answer is referring to cases where the input is invalid and specifically references catching the exception like so:

try {
    int x = ob.nextInt();
} catch (InputMismatchException e) {
    ob.next();
}

Which I seriously doubt has anything to do with your problem unless you're intentionally entering bad data.

Then there's this, the most likely culprit considering it's a potential infinite loop at first blush.

for (int i = 0; i < spstr.length; i++) {
    if (spstr[i].equals("")) {
        i--;
     } else {
         c = c + Integer.parseInt(spstr[i]);
     }
}

If i is 5 and spstr[i].equals("") returns true then i becomes 4, the else is skipped and i is incremented back to 5 ad infinitum.

Actually I forgot to use `trim()` while splitting the input, so I thought if there's any empty space it would count it as data and will store one less integer from input, so whenever it happens I decreased `i` so that i could have my input. But I was doing many mistakes in my code, as you said. Now got what I wanted to achieve. **Thank You so much for your help** @kab — Somendra Meena, Jun 04 '16 at 17:41

score 1 · Accepted Answer · answered Jun 03 '16 at 20:38

Simply use ob.nextLine() to ignore it. I fixed the code for you and it works as it should. Your code had several issues which I have mentioned.

import java.util.*;
class HelloWorld {
    public static void main(String args[]) {
        Scanner ob = new Scanner(System.in);
        int t = ob.nextInt(); 
        ob.nextLine();
        int a = 0, c = 0;
        for (int j = 0; j < t; j++)
        {
            a = 0; c = 0;
            String str = ob.nextLine();
            if(str.trim().length()>0){
            String [] spstr = str.trim().split("\\s+");
            try
            {
                for (int i=0 ; i<spstr.length ; i++)
                {
                    c = c + Integer.parseInt(spstr[i]);
                }
                System.out.println(c);
            } catch (NumberFormatException e) {
                System.out.println("Invalid Input");
            }
        }
    }
  }
}

if(spstr[i].equals("")) {i--;} is pointless and wrong logic in fact which will throw your program into an infinite loop. Simply trim the String and check if it is empty as I have done.
Do not simply catch Exception superclass. This is bad for debugging. The Exception raised here is NumberFormatException which you should catch.

In the catch block, you have not written the line `ob.nextLine();` and according to the post which I have linked in the question : **When a scanner throws an InputMismatchException, the scanner will not pass the token that caused the exception, so that it may be retrieved or skipped via some other method.** So how did the token passed to the next input ? — Somendra Meena, Jun 04 '16 at 17:32
Yes, I tried and it worked absolutely fine. Thanks for your help :) — Somendra Meena, Jun 04 '16 at 22:24

Unable to handle infinite loop when using Scanner for input

2 Answers2