Write a C++ program that creates n child threads, where each child thread prints a message into STDOUT with the assigned number by the parent thread. The range of the integer values assigned by the main thread to the child threads goes from 0 until n- 1. Your program must use synchronization mechanisms to guarantee that the child threads print the message in the opposite order they were created, starting with child threads with odd numbers and finishing with child threads with even numbers. Your program will receive from STDIN the number of child threads (nchildthreads). For nchildthreads = 5, the expected output is: I am Thread 3 I am Thread 1 I am Thread 4 I am Thread 2 I am Thread 0 NOTES 1. You must create n child threads (where n is the value that your program receives from STDIN). 2. The integer number assigned to each child thread has a value from 0 to n-1. 3. Zero is considered an even number. 4. You can add as many global variables as needed. 5. You can only use POSIX semaphores, pthreads mutex semaphore, or pthreads condition variables to achieve synchronization (A penalty of 100% will be applied to solutions using mechanisms other than the synchronization mechanisms listed before).
Write a C++ program that creates n child threads, where each child thread prints a message into STDOUT with the assigned number by the parent thread. The range of the integer values assigned by the main thread to the child threads goes from 0 until n- 1. Your program must use synchronization mechanisms to guarantee that the child threads print the message in the opposite order they were created, starting with child threads with odd numbers and finishing with child threads with even numbers. Your program will receive from STDIN the number of child threads (nchildthreads). For nchildthreads = 5, the expected output is: I am Thread 3 I am Thread 1 I am Thread 4 I am Thread 2 I am Thread 0 NOTES 1. You must create n child threads (where n is the value that your program receives from STDIN). 2. The integer number assigned to each child thread has a value from 0 to n-1. 3. Zero is considered an even number. 4. You can add as many global variables as needed. 5. You can only use POSIX semaphores, pthreads mutex semaphore, or pthreads condition variables to achieve synchronization (A penalty of 100% will be applied to solutions using mechanisms other than the synchronization mechanisms listed before).
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Need help completing C++ program
#include
#include
#include
#include
#include
static pthread_mutex_t bsem;
static pthread_cond_t waitTurn = PTHREAD_COND_INITIALIZER;
static int turn;
static int nthreads;
void *print_in_reverse_order_odd_then_even(void *void_ptr_argv)
{
// std::cout << "I am Thread " << /*variable identifier*/ << std::endl;
return NULL;
}
int main()
{
std::cin >> nthreads;
pthread_mutex_init(&bsem, NULL); // Initialize access to 1
pthread_t *tid= new pthread_t[nthreads];
int *threadNumber=new int[nthreads];
//HINT: this code determines the starting thread (thread with the highest odd number).
//You can erase this if statement if your solution does not need to know the starting child thread number.
if ((nthreads-1)%2!=0)
turn = nthreads -1;
else
turn = nthreads -2;
for(int i=0;i
Transcribed Image Text:4:32
Done
Scanned Documents
Write a C++ program that creates n child threads, where each child thread prints a message into
STDOUT with the assigned number by the parent thread. The range of the integer values assigned
by the main thread to the child threads goes from 0 until n - 1. Your program must use
synchronization mechanisms to guarantee that the child threads print the message in the
opposite order they were created, starting with child threads with odd numbers and finishing with
child threads with even numbers. Your program will receive from STDIN the number of child
threads (nchildthreads).
For nchildthreads = 5, the expected output is:
I am Thread 3
I am Thread 1
I am Thread 4
I am Thread 2
I am Thread e
NOTES
1. You must create n child threads (where n is the value that your program receives from
STDIN).
2. The integer number assigned to each child thread has a value from 0 to n-1.
3. Zero is considered an even number.
4. You can add as many global variables as needed.
5. You can only use POSIX semaphores, pthreads mutex semaphore, or pthreads
condition variables to achieve synchronization (A penalty of 100% will be applied to
solutions using mechanisms other than the synchronization mechanisms listed
before).
![main.cpp
#include <pthread.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <fcntl.h>
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2.
static pthread_mutex_t bsem;
static pthread_cond_t waitTurn = PTHREAD_COND_INITIALIZER;
static int turn;
static int nthreads;
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void *print_in_reverse_order_odd_then_even(void *void_ptr_argv)
13 {
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// std::cout << "I am Thread "
return NULL;
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<</*variable identifier*/ << std::endl;
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17
int main()
19 {
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std::cin >> nthreads;
pthread_mutex_init(&bsem, NULL); // Initialize access to 1
pthread t *tid= new pthread_t[nthreads];
int threadNumber=new int[nthreads]
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//HINT: this code determines the starting thread (thread with the highest odd number).
//You can erase this if statement if your solution does not need to know the starting
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if ((nthreads-1)%2!=0)
turn = nthreads -1;
else
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turn = nthreads -2;
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for(int i=0;i<nthreads; i++)
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// wait for the other threads to finish.
for (int i = 0; i < nthreads; i++)
pthread join(tid[i], NULL);
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delete [] threadNumber;
delete [] tid;
return e;
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Transcribed Image Text:main.cpp
#include <pthread.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <fcntl.h>
1
2.
static pthread_mutex_t bsem;
static pthread_cond_t waitTurn = PTHREAD_COND_INITIALIZER;
static int turn;
static int nthreads;
7
10
11
void *print_in_reverse_order_odd_then_even(void *void_ptr_argv)
13 {
12
// std::cout << "I am Thread "
return NULL;
14
<</*variable identifier*/ << std::endl;
15
16
17
int main()
19 {
18
std::cin >> nthreads;
pthread_mutex_init(&bsem, NULL); // Initialize access to 1
pthread t *tid= new pthread_t[nthreads];
int threadNumber=new int[nthreads]
20
21
22
23
24
//HINT: this code determines the starting thread (thread with the highest odd number).
//You can erase this if statement if your solution does not need to know the starting
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27
if ((nthreads-1)%2!=0)
turn = nthreads -1;
else
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31
turn = nthreads -2;
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33
for(int i=0;i<nthreads; i++)
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// wait for the other threads to finish.
for (int i = 0; i < nthreads; i++)
pthread join(tid[i], NULL);
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delete [] threadNumber;
delete [] tid;
return e;
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