Why in the above code it skips input string for n=1

Question

When I run the above code, it skips the first input string and without taking the input string, it prints hs.size() and jumps to n=2.But from n=2, the code runs absolutely fine.It takes the input string and adds in the hash set.

Why it skips the input string for n=1?? Please help.

public class ch 
{
    public static void main(String[] args) 
    {
        HashSet hs=new HashSet();
        String s;
        Scanner console=new Scanner(System.in);
        int n=console.nextInt();

        for(int i=0;i<n;i++)
        {   
            s=console.nextLine();
            hs.add(s);
            System.out.println(hs.size());
        }
    }
}

Tip: Save yourself some confusion and try to almost always use `nextLine` and do the `parseInt` after — OneCricketeer, Jun 08 '16 at 19:59

score 0 · Answer 1 · answered Jun 08 '16 at 20:00

It is because nextLine() directly after nextXXX() (XXX being any of Int, Float Double,etc) creates problem. nextXXX do not parse \n and nextLine() in such a case reads only \n left behind by nextXXX(). To resolve the issue, you can use an extra nextLine() just below nextInt() and your code will work just fine..

Why in the above code it skips input string for n=1

1 Answers1