unreachable code

Hell guys! This is my first post in this forum.The following code on compilation shows  :
Exception in thread "main" java.lang.Error: Unresolved compilation problem: Unreachable code
             at A.main(A.java:60)

Can anyone help me out what is the problem?
import java.util.ArrayList; import java.util.Scanner; class A { public static void main(String args[]) { Scanner sc=new Scanner(System.in); int t=sc.nextInt(); for(int l=1;l<=t;l++) { int j=0,flag=0,k=0; int N=sc.nextInt(); int a=sc.nextInt(); int b=sc.nextInt(); ArrayList<Integer> al=new ArrayList<Integer>(); for(int i=0;i<N;i++) { al.add(sc.nextInt()); } for(;;) { for(Integer list:al) { if(j==0 ||j%2==0) { if(list%a==0) { al.remove(list); flag=1; } else { flag=2; k=2; break; } } if(j%2!=0) { if(list%b==0) { al.remove(list); flag=2; } else { flag=1; k=2; break; } } if(k==2) { break; } } } if(flag==1) { System.out.println("Bob"); } else if(flag==2) { System.out.println("Alice"); } } } }

Don't mind me asking, but where do all the nextInt's come from ?
Without any context of what you're trying to do it's really gonna be hard to help.
Also why do you start an infinite loop, just in the hope of breaking out of it?
You also should have a look at The Java Naming Conventions

Inputs will come form my machine. Sir,if you think i write something terrible in my code please help me out as i am new to programming.

inputs normally do come from your machine, but at no point you ask the user to put something in
System.out.println("please enter a round number"); int nr = scanner.nextInt();

Actually the above code is for a  deciding a game winner.
Where , t=no of test cases
           N=sequence of integers
           a=no selected by Bob
           b=no selected by Alice
Game will start from Bob and then Alice, alternate turns will occur in which both have to remove multiples of  no selected by them one by one . if  one is  not able to remove  multiple of their selected no the other will win.

for(;;)            {                for(Integer list:al)                {
The outer loop of this pair is an infinite loop. It can only end by using a break, return or throw statement. You do have some break statements, but these are all located inside the inner loop. As a result, these break out of the inner loop. That means the outer loop will still continue forever.

If you want to break out of the outer loop using a break statement in the inner loop, you need to use a labeled break.

Shubham Rai wrote:The following code on compilation shows  :
Exception in thread "main" java.lang.Error: Unresolved compilation problem: Unreachable code
             at A.main(A.java:60)

The compiler is complaining because for(;;) is an infinite loop and the code execution can not progress past it.

for(;;) { ... } if(flag==1) {    System.out.println("Bob"); } else if(flag==2) {    System.out.println("Alice"); }

Thank You ! you solve my problem .I found my mistake and problem resolved.

Daniel Demesmaecker wrote:
Also why do you start an infinite loop, just in the HOPE of breaking out of it?

I actually already maded that remark

Well  after one player's turn i have to start iterating the arrayList again from 0 index to check whether any multiple of no selected by the current player is present or not in list. and so on this process will repeat as many times, until winner will not be decided .
To overcome this problem i think let's use the infinite loop and when winner will be decided ,loop will be broke by break.

an while or do whileloop using a boolean would be cleaner...