Informatica Interview Question


Country: India
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
3
of 5 vote

class Q {
    void A1()
    { 
        System.out.println("In A1");
    }
    void B1()
    { 
        System.out.println("In B1");
    }
    void A2()
    { 
        System.out.println("In A2");
    }
    void B2()
    { 
        System.out.println("In B2");
    }    
    
    boolean l1 = false;
    boolean l2 = false;
    boolean l3 = false;    
    
    synchronized void Thread1()
    {
        A1();
        
        l1 = true;
        notify();        
        
        while(!l2)
        {
            try
            {
                wait();
            }
            catch(InterruptedException e)
            {
                System.out.println("Interrupted exception caught\n");
            }
        }
                
        A2();
                
        l3 = true;
        notify();
    }        
    
    synchronized void Thread2()
    {
        while(!l1)
        {
            try
            {
                wait();
            }
            catch(InterruptedException e)
            {
                System.out.println("Interrupted exception caught\n");
            }      
        }               
               
        B1();
        
        l2=true;        
        notify();
                
        
        while(!l3)
        {
            try
            {
                wait();
            }
            catch(InterruptedException e)
            {
                System.out.println("Interrupted exception caught\n");
            }      
        }         
        B2();
    }     
}


class Sync implements Runnable {
    Sync(Q q,int tid)
    {
        this.q = q;
        this.tid = tid;
        thr = new Thread(this);
        thr.start();        
    }
    
    public void run()
    {
        if (tid == 1)
        {
            q.Thread1();                             
        }
        if( tid == 2)
        {           
            q.Thread2();
        }
    }
    Q q; 
    int tid;
    Thread thr;
}
class Srujana {
    public static void main(String args[])
    {
        Q q = new Q();
        new Sync(q,1);
        new Sync(q,2);
    }
}

- srujana October 01, 2012 | Flag Reply
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1
of 1 vote

what will you do if there are more than A1,A2.. if this keep growing?

- Anonymous March 01, 2013 | Flag
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2
of 2 vote

public class SyncThread {
/**
* Please see the below code its very simple. 
*
*/
	final Object o = new Object();
	volatile int count;   // counter 
	volatile boolean isT1started = false;
	final String ResourceA1="A1", ResourceA2 ="A2", ResourceB1="B1", ResourceB2="B2";
	
	// Thread T1

	Runnable T1 = new Runnable() {
		@Override
		public void run() {
   		while(count<5){
			synchronized (o) {
				try {
					isT1started = true;
					System.out.println(ResourceA1);
					o.notify();
					o.wait();
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
			
			synchronized (o) {
			try {
				System.out.println(ResourceA2);
				o.notify();
				o.wait();
				} catch (InterruptedException e) {
					e.printStackTrace();
				}
			}
		}
		}
	};
	
	// Thread T2
	Runnable T2 = new Runnable() {
		@Override
		public void run() {
		while(count<5){
			try {
				synchronized (o) {
					System.out.println(ResourceB1);
					o.notify();
					o.wait();
				}

				synchronized (o) {
					System.out.println(ResourceB2);
					System.out.println("============count completed:"+  (++count));
			o.notify();
			o.wait();
				}
			} catch (Exception e) {
			e.printStackTrace();
			}
		}
		}
	};
	
	void startThread(){
		int i=1;
		Thread t1 = new Thread(T1);
		Thread t2 = new Thread(T2);
		t1.start();

// makes sure that Thread t1 is started before t2 
	while(!isT1started){
			try {
				Thread.sleep(100);
			} catch (InterruptedException e) {
				e.printStackTrace();
			}
		}
// Now start thread t2
		t2.start();
		
		
	}
	
	public static void main(String[] args) {
	new SyncThread().startThread();
	}
}

- san4net October 02, 2012 | Flag Reply
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0
of 0 votes

You can fix it simply by moving waiting into Thread 2. Something like

Runnable T2 = new Runnable()
		{
			@Override
			public void run()
			{
				while (count < 5)
				{
					try
					{
						//Ensure T2 Runs always after T1
						while (isT1started)
						{
							System.out.println("T1 has not run once so T2 Starts waiting");
							Thread.sleep(500);
						}

- coool.aman June 24, 2013 | Flag
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1
of 1 vote

Following is my code using Events kernel objects in C++.

HANDLE h[4] = {0};  


void A1(void)
{
	std::cout<< "A1\n"; 
} 

void A2(void)
{
	std::cout<< "A2\n"; 
}

void B1(void)
{
	std::cout<< "B1\n"; 
}

void B2(void)
{
	std::cout<< "B2\n"; 
}


DWORD WINAPI Thread1(LPVOID t1)
{

	WaitForSingleObject(h[0],INFINITE);
	A1(); 
	SetEvent(h[2]);
	WaitForSingleObject(h[1],INFINITE);
	A2(); 
	SetEvent(h[3]);
	return 0;

}


DWORD WINAPI Thread2(LPVOID t2)  
{
	WaitForSingleObject(h[2],INFINITE);
	B1(); 
	SetEvent(h[1]);
	WaitForSingleObject(h[3],INFINITE);
	B2(); 
	return 0;
}


int main(int argc, char* argv[])
{
	int  temp=0;
	int i=0;
	HANDLE hThread[2] = {0}; 

	h[0] = CreateEvent(NULL,FALSE,TRUE,NULL);  // For A1. Signaled State.
	h[1] = CreateEvent(NULL,FALSE,FALSE,NULL); // For A2. Non-signaled state.
	h[2] = CreateEvent(NULL,FALSE,FALSE,NULL); // For B1. Non-signaled state. 
	h[3] = CreateEvent(NULL,FALSE,FALSE,NULL); // For B2. Non-signaled state.  
	
	hThread[0] = CreateThread(NULL, 0, Thread1, &temp, 0, NULL);
	hThread[1] = CreateThread(NULL, 0, Thread2, &temp, 0, NULL);


	WaitForMultipleObjects(2,hThread,true,INFINITE);
	WaitForMultipleObjects(2,h,true,INFINITE);


	while(i<2)
	 CloseHandle(hThread[i++]);
	
	i=0;
	while(i<4)
	 CloseHandle(h[i++]);

    getchar();
	return 0; 
}

- rahul.s July 24, 2013 | Flag Reply
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0
of 0 vote

ammm .... this code is wrong ... as you are assuming that Thread 1 should be started before Thread 2 ....

What if OS preempts the thread1 just after start of execution but before reaching any executable statement.......and what if OS schedules thread 2 ecen if it is started after Thread 2 ??? .... In this case your program will fail..... right ???

The basic principle about multithreading is .....,Done make any assumptions about the sequence in which threads will execute .... Because you never know how OS will schedule them at run time .....

- Prafulla Vedante October 07, 2012 | Flag Reply
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0
of 0 vote

Here is an implementation using a ConcurrentLinkedQueue, let me know what you folks think

public class SequencingUsingBoundedBlockingQueue {
	private static class Resource {
		public final String name;
		public Resource(String name) {
			this.name = name;
		}
		public void useResource() {
			System.out.println("Resource " + name + " used.");
		}
	}
	private final Queue<Resource> queue;
	public SequencingUsingBoundedBlockingQueue() {
		queue = new ConcurrentLinkedQueue<Resource>();
		queue.add(new Resource("A"));
		queue.add(new Resource("B"));
		queue.add(new Resource("C"));
		queue.add(new Resource("D"));
	}
	private static class Actor implements Runnable {
		private final Set<String> resources;
		private final Queue<Resource> queue;
		public Actor(Set<String> resources, Queue<Resource> queue) {
			this.resources = resources;
			this.queue = queue;
		}

		@Override
		public void run() {
			while (!queue.isEmpty()) {
				Resource r = queue.peek();
				if (r == null) {
					break;
				}
				if (resources.contains(r.name)) {
					queue.remove().useResource();
				}
			}
		}	
	}
	
	public static void main(String args[]) throws Exception {
		SequencingUsingBoundedBlockingQueue problem = new SequencingUsingBoundedBlockingQueue();
		Set<String> setA = new HashSet<String>();
		setA.add("A"); setA.add("B");
		Set<String> setB = new HashSet<String>();
		setA.add("C"); setA.add("D");
		(new Thread(new Actor(setB, problem.queue))).start();
		(new Thread(new Actor(setA, problem.queue))).start();

- Anonymous October 10, 2012 | Flag Reply
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0
of 0 votes

Incorrectly named the class, must have named it SequencingUsingConcurrentLinkedQueue, sorry about that. Btw this code compiles and runs fine, try it out and let me know

- Anonymous October 10, 2012 | Flag
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0
of 2 vote

Guys how about this , controlling the access using 4 mutexes.

#include <pthread.h>
#include <iostream>
using namespace std;

pthread_mutex_t m;
pthread_mutex_t a1,a2,b1,b2;

extern "C" void *fun1(void *)
{
        pthread_mutex_lock(&a1);
        cout<<endl<<"A1"<<endl;
        pthread_mutex_unlock(&a1);

        pthread_mutex_lock(&b2);
        pthread_mutex_lock(&b1);
        cout<<endl<<"A2"<<endl;
        pthread_mutex_unlock(&b1);
        pthread_mutex_unlock(&b2);
        return NULL;

}
extern "C" void *fun2(void *)
{
        pthread_mutex_lock(&b1);
        pthread_mutex_lock(&a1);
        cout<<endl<<"B1"<<endl;
        pthread_mutex_unlock(&a1);
        pthread_mutex_unlock(&b1);


        pthread_mutex_lock(&b2);
        cout<<endl<<"B2"<<endl;
        pthread_mutex_unlock(&b2);
        return NULL;
}

main()
{
        pthread_t thread_id1,thread_id2;

        pthread_create(&thread_id1,NULL,&fun1,NULL);
        pthread_create(&thread_id2,NULL,&fun2,NULL);
		
        pthread_join(thread_id1,NULL);
        pthread_join(thread_id2,NULL);
}

- BJ February 15, 2013 | Flag Reply
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0
of 0 votes

can you explain your code ? It sounds like deadlock

- keyurpatel80 June 02, 2013 | Flag
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0
of 0 vote

interface TaskA {
    public void execute();
}

class TaskA1 implements TaskA {
    public void execute() { System.out.println("TaskA1"); }
}

class TaskA2 implements TaskA {
    public void execute() { System.out.println("TaskA2"); }
}

interface TaskB {
    public void execute();
}

class TaskB1 implements TaskB {
    public void execute() { System.out.println("TaskB1"); }
}

class TaskB2 implements TaskB {
    public void execute() { System.out.println("TaskB2"); }
}

class TaskManager {
    private LinkedList<TaskA> taskAQ = new LinkedList<TaskA>();
    private LinkedList<TaskB> taskBQ = new LinkedList<TaskB>();

    boolean flag = true;

    public synchronized void addTaskA(TaskA a) {
        taskAQ.add(a);
        notifyAll();
    }
    public synchronized void addTaskB(TaskB b) {
        taskBQ.add(b);
        notifyAll();
    }

    public synchronized void doTaskA() {
        try {
            while(taskAQ.size() == 0)
                wait();

            while(flag == false) //supposed to be taskb's turn
                wait();
        }catch(InterruptedException e) {}

        TaskA a = taskAQ.poll();
        a.execute();
        flag = false;
        notify();
    }

    public synchronized void doTaskB() {
        try {
            while(taskBQ.size() == 0)
                wait();
            
            while(flag == true)
                wait();
        }catch(InterruptedException e) {}

        TaskB b = taskBQ.poll();
        b.execute();
        flag = true;
        notify();
    }
}

class ThreadA extends Thread {
    private TaskManager tm;

    public ThreadA (TaskManager t) { tm = t;}
    public void run() {
        while(true)
            tm.doTaskA();
    }
}

class ThreadB extends Thread {
    private TaskManager tm;

    public ThreadB (TaskManager t) { tm = t;}
    public void run() {
        while(true)
            tm.doTaskB();
    }
}

class ThreadP extends Thread {
    private TaskManager tm;

    public ThreadP(TaskManager t) { tm = t;}

    public void run() {
        
        TaskA a1 = new TaskA1();
        TaskA a2 = new TaskA2();

        TaskB b1 = new TaskB1();
        TaskB b2 = new TaskB2();

        tm.addTaskA(a1);
        tm.addTaskA(a2);
        tm.addTaskA(new TaskA() {
                public void execute() { System.out.println("Anonymous TaskA3");}
            });

        tm.addTaskB(b1);
        tm.addTaskB(b2);
        tm.addTaskB(new TaskB() {
                public void execute() { System.out.println("Anonymous TaskB3");}
            });

    }
}
public class test {
    public static void main(String args[]) {

        TaskManager tm = new TaskManager();

        ThreadA a = new ThreadA(tm);
        ThreadB b = new ThreadB(tm);
        ThreadP p = new ThreadP(tm);
        

        a.start();
        b.start();
        p.start();
   }
}

- s.kalyanasundaram March 01, 2013 | Flag Reply
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0
of 0 vote

public class Test2 implements Runnable{

String first = "";
String second = "";
static Integer lock = 1;


public Test2(String first, String second) {
this.first = first;
this.second = second;
}


@Override
public void run() {
synchronized (lock) {
System.out.println(first);
lock.notify();
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println(second);
lock.notify();
}
}





public static void main(String[] args) {

Thread t1 = new Thread(new Test2("A1", "A2"));
Thread t2 = new Thread(new Test2("B1", "B2"));

t1.start();
t2.start();

}

}

- coder June 05, 2013 | Flag Reply
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0
of 0 vote

package informatica;

public class Multi{
public static Boolean mutex1 =false;
public static final Object obj= new Object();
//public static Boolean mutex2 =false;

class T1 extends Thread{

public void A1(){
System.out.println("IN A1");
}
public void A2(){
System.out.println("IN A2");
}

public void run(){

synchronized (obj) {
if(mutex1==false){
A1();
mutex1=true;
//System.out.println("Notifying");
obj.notify();
}else{
//System.out.println("Else of A1");
try {
obj.wait();
if(mutex1==false){
A1();
mutex1=true;
}
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}

synchronized (obj) {
if(mutex1==false){
A2();
mutex1=true;
//System.out.println("Notifying");
obj.notify();
}else{
//System.out.println("Else of A1");
try {
obj.wait();
if(mutex1==false){
A2();
mutex1=true;
obj.notify();
}
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}

/*
if(mutex1==false){
A1();
mutex1=true;
}
while(true){
if(mutex1==false){
A2();
mutex1=true;
break;
}else{
try{
Thread.sleep(100);
}catch(Exception e){
e.printStackTrace();
}
}
}
*/
}
}
class T2 extends Thread{

public void B1(){
System.out.println("IN B1");
}
public void B2(){
System.out.println("IN B2");
}
public void run(){

synchronized (obj) {
//System.out.println("Sync of B1");
if(mutex1==true){
B1();
mutex1=false;
obj.notify();
}else{
//System.out.println("Else of B1");
try {
obj.wait();
if(mutex1==true){
B1();
mutex1=false;
obj.notify();
}
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}

}
}

synchronized (obj) {
//System.out.println("Sync of B1");
if(mutex1==true){
B2();
mutex1=false;
obj.notify();
}else{
//System.out.println("Else of B1");
try {
obj.wait();
if(mutex1==true){
B2();
mutex1=false;
}
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}

}
}

/*
while(true){
if(mutex1==true){
B1();
mutex1=false;
break;
}else{
try{
Thread.sleep(100);
}catch(Exception e){
e.printStackTrace();
}
}
}
while(true){
if(mutex1==true){
B2();
mutex1=false;
break;
}else{
try{
Thread.sleep(100);
}catch(Exception e){
e.printStackTrace();
}
}
}
*/
}


}


public static void main(String[] args) {
Multi m = new Multi();
T1 t1= m.new T1();
T2 t2= m.new T2();

t2.start();
t1.start();
}

}

- Anonymous June 18, 2013 | Flag Reply
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0
of 0 vote

import java.util.concurrent.Semaphore;

public class ThreadAlternate {

public static void main(String[] args) {
String[] s1 = {"A1","A2","A3"};
String[] s2 = {"B1","B2","B3"};
MyThread t1 = new MyThread(s1, true);
MyThread t2 = new MyThread(s2, false);
t1.start();
t2.start();
}

}

class MyThread extends Thread{

static Semaphore s1 = new Semaphore(1);
static Semaphore s2 = new Semaphore(1);
boolean flag = false;
static{
try {
s2.acquire();
} catch (InterruptedException e) {
e.printStackTrace();
}
}

String[] arr;
public MyThread(String[] strArray, boolean flag) {
this.arr = strArray;
this.flag = flag;
}

public void run() {
for (int i = 0; i < arr.length; i++) {
if(flag){
try {
s1.acquire();
} catch (InterruptedException e) {
e.printStackTrace();
}
System.err.println(arr[i]);
s2.release();
}else{
try {
s2.acquire();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.err.println(arr[i]);
s1.release();
}
}

}
}

- Anonymous July 25, 2013 | Flag Reply
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0
of 0 vote

import java.util.concurrent.Semaphore;

public class ThreadAlternate {

	public static void main(String[] args) {
		String[] s1 = {"A1","A2","A3"};
		String[] s2 = {"B1","B2","B3"};
		MyThread t1 = new MyThread(s1, true);
		MyThread t2 = new MyThread(s2, false);
		t1.start();
		t2.start();
	}

}

class MyThread extends Thread{

	static Semaphore s1 = new Semaphore(1);
	static Semaphore s2 = new Semaphore(1);
	boolean flag = false;
	static{
		try {
			s2.acquire();
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
	}

	String[] arr;
	public MyThread(String[] strArray, boolean flag) {
		this.arr = strArray;
		this.flag = flag;
	}

	public void run() {
		for (int i = 0; i < arr.length; i++) {
			if(flag){
				try {
					s1.acquire();
				} catch (InterruptedException e) {
					e.printStackTrace();
				}
				System.err.println(arr[i]);
				s2.release();
			}else{
				try {
					s2.acquire();
				} catch (InterruptedException e) {
					// TODO Auto-generated catch block
					e.printStackTrace();
				}
				System.err.println(arr[i]);
				s1.release();
			}
		}

	}
}

- Anonymous July 25, 2013 | Flag Reply
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0
of 0 vote

let X and Y be two binary semaphores where X = 1 and Y = 0

Thread1						Thread2
{							{
	wait(X)						wait(Y)
	print(A1)						print(B1)
	signal(Y)					signal(X)

	wait(X)						wait(Y)
	print(A2)						print(B2)
	signal(Y)					signal(X)
}							}

Tell if the above logic will work or not.

- Vijay November 25, 2013 | Flag Reply
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0
of 0 votes

Your logic wont work. According to ur logic,
A1 B1 A1 can be printed, as after printing B1 you are doing signal(X).
if Thread 1 is scheduled just after this, A1 can again be printed.

To maintain the order, you need to use 4 mutuxes
m_A1 = 1, m_A2 = 0, m_B1 = 0, m_B2 = 0;

thread1() {

wait(m_A1);
print A1
signal(m_B1)
wait(m_A2)
print A2
signal(m_B2)
}

thread2{
wait(m_B1)
print B1
signal(m_A2)

wait(m_B2)
print B2
signal(m_A1)
}

- vivekh August 10, 2015 | Flag
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0
of 0 vote

We can use conditional variables for this (pthread_cond_wait and pthread_cond_signal).
B1 should wait on a signal from A1.
A2 should wait on a signal from B1
and so on.
So we need 3 conditional variables for this and a mutex variable

- Sumit Das February 24, 2014 | Flag Reply
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0
of 0 vote

package com.home;

/**
* App
*
*/
public class App
{
public static String CLASS_NAME = "App :";
public static void main( String[] args )
{
Thread T1 = new Thread(r1);
Thread T2 = new Thread(r1);

try{
T1.start();
T2.start();

T2.join();

} catch(Exception e){
e.printStackTrace();
System.out.println("error :" +e.getMessage());
}

}

static volatile boolean t2Start = false;
static Runnable r1 = new Runnable(){

public void run() {
try {
System.out.println("t2Start:" + t2Start);
if(!t2Start){
synchronized (this) {
System.out.println(Thread.currentThread().getName());
System.out.println("A1");
t2Start = true;
}
synchronized (this) {
System.out.println(Thread.currentThread().getName());
System.out.println("B1");

}
}

if(t2Start){
synchronized (this) {
System.out.println(Thread.currentThread().getName());
System.out.println("A2");
}


synchronized (this) {
System.out.println(Thread.currentThread().getName());
System.out.println("B2");
}

}

} catch (Exception e) {
e.printStackTrace();
System.out.println("ERRORA: " + e.getMessage());
}
}
};






}

- Anonymous November 21, 2014 | Flag Reply
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0
of 0 vote

public class Example
{

    public static void main(String[] args) throws InterruptedException
    {
        final Object o = new Object();
        Thread t1 = new Thread(new Runnable()
        {

            @Override
            public void run()
            {
                synchronized (o)
                {
                    System.out.println("A1");
                    try
                    {
                        o.wait();
                    } catch (InterruptedException e)
                    {
                        e.printStackTrace();
                    }

                    System.out.println("A2");
                    o.notify();
                }

            }
        });
        Thread t2 = new Thread(new Runnable()
        {

            @Override
            public void run()
            {
                try
                {
                    Thread.sleep(100);
                } catch (InterruptedException e)
                {
                    e.printStackTrace();
                }
                synchronized (o)
                {
                    System.out.println("B1");
                    o.notify();
                    try
                    {
                        o.wait();
                    } catch (InterruptedException e)
                    {
                        e.printStackTrace();
                    }

                    System.out.println("B2");
                }
            }
        });
        t1.start();
        t2.start();

        t1.join();
        t2.join();
    }
}

- Naveen March 01, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <iostream>
#include <thread>
#include <mutex>
#include <vector>
#include <condition_variable>

using namespace std;

class Semaphore {
private:
	mutex mtx;
	condition_variable cv;
	int counter;
public:
	Semaphore(int value) {
		counter = value;
	}
	void wait() {
		unique_lock <mutex> lck(mtx);
		cv.wait(lck, [&] {return counter > 0; });	//what does & do??
		counter--;
	}
	void signal() {
		unique_lock <mutex> lck(mtx);
		counter++;
		cv.notify_one();
	}
};

Semaphore tA(1), tB(0);
void threadA() {
	tA.wait();
	cout << "A1-";
	tB.signal();
	tA.wait();
	cout << "A2-";
	tB.signal();
}

void threadB() {
	tB.wait();
	cout << "B1-";
	tA.signal();
	tB.wait();
	cout << "B2";
}

int main() {
	thread t1(threadA);
	thread t2(threadB);
	getchar();
	return 0;
}

- Anonymous March 05, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <iostream>
#include <thread>
#include <mutex>
#include <vector>
#include <condition_variable>

using namespace std;

class Semaphore {
private:
	mutex mtx;
	condition_variable cv;
	int counter;
public:
	Semaphore(int value) {
		counter = value;
	}
	void wait() {
		unique_lock <mutex> lck(mtx);
		cv.wait(lck, [&] {return counter > 0; });	//what does & do??
		counter--;
	}
	void signal() {
		unique_lock <mutex> lck(mtx);
		counter++;
		cv.notify_one();
	}
};

Semaphore tA(1), tB(0);
void threadA() {
	tA.wait();
	cout << "A1-";
	tB.signal();
	tA.wait();
	cout << "A2-";
	tB.signal();
}

void threadB() {
	tB.wait();
	cout << "B1-";
	tA.signal();
	tB.wait();
	cout << "B2";
}

int main() {
	thread t1(threadA);
	thread t2(threadB);
	t1.join();
	t2.join();
	getchar();
	return 0;
}

- ngahlot March 05, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

{{

public class SequencePrinter {

public static void main(String args[])
{
Threads T1 = new Threads("a1","a2");
Threads T2 = new Threads("b1","b2");
T1.start();
T2.start();
}

}
class Threads extends Thread
{
String s1 = null;
String s2 = null;
int count = 0;
Threads(String s1, String s2)
{
this.s1 = s1;
this.s2 = s2;
this.count = 0;
}
public void run()
{
if(count == 0)
{
System.out.println(this.s1);
count++;
this.yield();
}
if(count == 1)
{
System.out.println(this.s2);
}
}
}




}}

- TonyStark March 19, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote
public class ThreadExample1 {{{ String s1="A1"; String s2="A2"; String s3="B1"; String s4="B2"; Object lock1= new Object(); Object lock2=new Object(); boolean thread1Started=false; boolean thread2Started=false; class Thread1 extends Thread{{{ @Override public void run() {{{ synchronized (lock1) {{{ System.out.println(s1); thread1Started=true; while (!thread2Started) {{{ }}}; synchronized (lock2) {{{ System.out.println(s2); lock2.notifyAll(); }}} }}} }}} }}} class Thread2 extends Thread{{{ @Override public void run() {{{ while (!thread1Started) {{{ }}}; synchronized (lock2) {{{ System.out.println(s3); thread2Started=true; try {{{ lock2.wait(); }}} catch (InterruptedException e) {{{ e.printStackTrace(); }}} synchronized (lock1) {{{ System.out.println(s4); }}} }}} }}} }}} public static void main(String[] args) {{{ int i=0; while (i<=10) {{{ ThreadExample1 threadExample1=new ThreadExample1(); ThreadExample1.Thread1 thread1=threadExample1.new Thread1(); ThreadExample1.Thread2 thread2=threadExample1.new Thread2(); System.out.println("Iteration NO::"+i++); thread2.start(); thread1.start(); }}} }}} }}} - prasenjit235 July 06, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
Comment hidden because of low score. Click to expand.
0
of 0 votes

This is so wrong man.

- Teddy October 02, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

lol

- Noobie October 02, 2012 | Flag
Comment hidden because of low score. Click to expand.


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