Why does the previous program print what you selected?

Why does the previous program print what you selected?

Why does the previous program print what you selected?

Question 1 (1 point) What does the following program print? Note important sections in bold. You may assume executor.submit() is first-in-first-out. import java.util.concurrent.Callable; import java.util.ArrayList; import java.util.List; import java.util.concurrent. ExecutionException; import java.util.concurrent. ExecutorService; import java.util.concurrent. Executors; import java.util.concurrent. Future; public class FutureExample implements Callable { private int value; FutureExample(int v) { value = V; } @Override public Integer call() throws Exception { Thread.sleep((long) (Math.random() * 10)); return value; } public static void main(String args[]) { Executor Service executor = Executors.newFixedThreadPool(10) List> list = new ArrayList> for(int i=0; i callable = new FutureExample(i); Future future = executor.submit(callable); list.add(future); } for(Future fut : list) { try { System.out.println(fut.get()); } catch (InterruptedException | ExecutionException e) { e.printStackTrace(); } } executor.shutdown(); } } Integers from 0 to 99, in order. Integers from 0 to 99, but not in order Random integers, without clear upper bounds None of these. Question 3 (1 point) What does the following program print? Note important sections in bold. You may assume executor.submit() is first-in-first-out. import java.util.concurrent. Callable; import java.util.ArrayList; import java.util.List; import java.util.concurrent. ExecutionException; import java.util.concurrent. ExecutorService; import java.util.concurrent. Executors; import java.util.concurrent. Future; import java.util.concurrent.atomic.*; public class FutureExample implements Callable { Atomic Integer v = new AtomicInteger(0); @Override public Integer call() throws Exception { Thread.sleep( (long) (Math.random() * 10)); return v.getAndIncrement(); } public static void main(String args[]) { ExecutorService executor = Executors.newFixedThreadPool(1 0); List> list = new ArrayList> Callable callable = new FutureExample(); for(int i=0; i future = executor.submit(callable); list.add(future); } for(Future fut : list) { try { System.out.println(fut.get()); } catch (InterruptedException | ExecutionException e) { e.printStackTrace(); } } executor.shutdown(); } } Integers from 0 to 99, in order. Integers from 0 to 99, but not in order Random integers, without clear upper bounds None of these. Question 5 (1 point) What does the following program print? Note important sections in bold. You may assume executor.submit() is first-in-first-out. import java.util.concurrent. Callable; import java.util.ArrayList; import java.util.List; import java.util.concurrent. ExecutionException; import java.util.concurrent. ExecutorService; import java.util.concurrent. Executors; import java.util.concurrent. Future; import java.util.concurrent.atomic.*; public class FutureExample implements Callable { Atomic Integer v = new AtomicInteger(0); @Override public Integer call() throws Exception { Thread.sleep((long) (Math.random() * 10)); return v.getAndIncrement(); } public static void main(String args[]) { ExecutorService executor = Executors.newFixedThreadPool(1 ); // 1, not 10 List> list = new ArrayList> Callable callable = new FutureExample(); for(int i=0; i future = executor.submit(callable); list.add(future); } for(Future fut : list) { try { System.out.println(fut.get()); } catch (InterruptedException | ExecutionException e) { e.printStackTrace(); } } executor. shutdown(); } } Integers from 0 to 99, in order. Integers from 0 to 99, but not in order Random integers, without clear upper bounds None of these