Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people, just like you, are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions

Single writer , multiple reader in Real Time code

Posted on 2009-05-03
Last Modified: 2013-11-25
Dear experts,

In real time programming, it is common that we need to have concurrency procedure that enable only 1 single writer OR multiple concurrent reader to one shared variable / memory at one time.
Is there anyone can share common high level design (in terms of how to handle the concurrency , locking, queuing) for this (not specific to any language)? Or any references?

Thanks so much.
Question by:isuhendro
  • 3
  • 3
  • 2
  • +2
LVL 86

Expert Comment

ID: 24290279
There are high level concepts such as mutexes, locking, semaphores etc but in java, look at the java.util.concurrent package classes
LVL 40

Assisted Solution

evilrix earned 50 total points
ID: 24290473
I don't know if it helps but I wrote the code example below for another question but it does exhibit the qualities you discuss. It is an example of (a simple) solution to a producer/consumer pattern. The producer (the main thread and single writer) writes requests to a queue (your shared variable) and the threads in the thread pool (the multiple readers) read and process the requests. The code using mutual exclusion (mutex) to ensure the variable is only being access/modified by one thread at a time. A single event is used to signal the thread pool to dispatch a thread to process a newly queue request. Additionally, another even is used to signal the thread pool to terminate.

Since the code exhibits the requirements you discussed I figured I'd post it so you could take a look and see how there constructs have been used to help implement this solution. The same principles can, obviously, been applied to your more specific requirement.

I hope this helps.
include <windows.h>
#include <process.h>
#include <deque>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
// Typedefs just to aid readability
typedef std::deque<std::string> string_queue_t;
typedef std::vector<HANDLE> handle_vector_t;
// Thread data
struct thread_data
        string_queue_t string_queue;
        HANDLE mtx;
        HANDLE evts[2]; // 0 Quit, 1 Process
// A thread
void thread_proc(void * pv)
        // If we have no thread data terminate
        if(0 == pv) return;
        // Get the thread data
        thread_data & td = *reinterpret_cast<thread_data *>(pv);
        bool bQuit = 0;
                // Wait for either a job or signal to quit
                DWORD result = WaitForMultipleObjects(2, td.evts, false, INFINITE);
                // Get exclusive access to the queue
                WaitForSingleObject(td.mtx, INFINITE);
                // Take a copy of the job in the queue
                // NB. working with a copy means we can free up access to the queue immediately
                std::string s;
                if(td.string_queue.size() > 0)
                        s = td.string_queue.front();
                        // The queue is empty so if we were signaled to quit then we can do so now
                        bQuit = ((result - WAIT_OBJECT_0) == 0);
                // Release exclusinve lock on queue
                // Process what we got out of the queue
                // NB. Since this is a copy the queue has been released from exclusive access
                        // This output is thread safe but may end up mixing with other threads
                        // that's ok it's just for example purposes.
                        std::cout << "tid: " << GetCurrentThreadId() << " - " << s << std::endl;
int main()
        // Create out cross-thread object
        thread_data td;
        // Mutex to sync access to the queue
        td.mtx = CreateMutex(0, false, 0);
        // Event used to signal threads to terminate
        td.evts[0] = CreateEvent(0, true, false, 0);
        // Event used to signal the thread pool a job is ready
        td.evts[1] = CreateEvent(0, false, false, 0);
        // A vector that will represent the thread pool
        handle_vector_t handle_vector;
        // Create thread pool
        for(int i = 0 ; i < 10 ; ++i)
                HANDLE handle = reinterpret_cast<HANDLE>(_beginthread(thread_proc, 0, &td));
                if(handle > 0)
        // Queue jobs for processing by thread pool
        for(int j = 0 ; j < 9999 ; ++j)
                std::stringstream ss;
                ss << "data " << j;
                // Add to queue
                WaitForSingleObject(td.mtx, INFINITE);
                // Signal the thread pool
        // Signal threads to terminate when there are no more jobs
        // wait for threads to terminate
        WaitForMultipleObjects(3, &handle_vector[0], true, INFINITE);

Open in new window


Author Comment

ID: 24292399
hi CEHJ, thanks for your advice.

However, I need specific information about multiple read & single writer. Let's say i have 3 threads of reader, and 1 thread of writer, I would like that whenever writer in action then no other thread would be able to perform. But whenever one reader in action, it could be that all of them (3 reader) do the reading at the same time.

Hi evilrixD, thank you for sharing the code.

If I don't understand wrongly your code is producer and consumer scenario.. however i could not see the fact that producer is prohibit to produce message when consumer is reading, or consumer is allowed to consume message whenever other consumer is in action.

I think the code required would need a conditioned variable to signify the current mode is R/W. However i am still confused the fact that we need to have multiple reader, against the fact that R/W cannot be run together.

Free Tool: Path Explorer

An intuitive utility to help find the CSS path to UI elements on a webpage. These paths are used frequently in a variety of front-end development and QA automation tasks.

One of a set of tools we're offering as a way of saying thank you for being a part of the community.

LVL 92

Accepted Solution

objects earned 75 total points
ID: 24292463
LVL 92

Expert Comment

ID: 24292472

Author Comment

ID: 24292691
hi Objects, thanks for the good out of the box answer.

i just found from the web a good and sounds of working algorithm here ,
the key is reader will only try to acquire "lock" if and only if it is the "first" active reader. the rest of the reader can do "reading" activity without any "lock"

// number of readers
int readcount = 0;
// mutual exclusion to readcount
Semaphore mutex = 1;
// exclusive writer or reading
Semaphore w_or_r = 1;
writer {
  wait(w_or_r); // lock out readers
  signal(w_or_r); // up for grabs
reader {
  wait(mutex); // lock readcount
  readcount += 1; // one more reader
  if (readcount == 1)
    wait(w_or_r); // synch w/ writers
  signal(mutex); // unlock readcount
  wait(mutex); // lock readcount
  readcount -= 1; // one less reader
  if (readcount == 0)
    signal(w_or_r); // up for grabs
  signal(mutex); // unlock readcount}

Open in new window

LVL 92

Expert Comment

ID: 24292697
LVL 40

Expert Comment

ID: 24293400
>> i could not see the fact that producer is prohibit to produce message when consumer is reading, or consumer is allowed to consume message whenever other consumer is in action.

Protection is provided via the mutex. See both the producer and consumer try to take ownership of the mutex (lines 38 and 103). Only one thread will gain ownership and the other will block until ownership is released by the other thread. This is a typical mechanism used to provide mutual exclusion semantics. You would avoid using a variable for a number of reasons...

1. Unless the variable is volatile changes in one thread may not be apparent in the others (due to CPU caching and compiler optimization)

2. Unless the variable type can be read/written/tested in one instruction you have the strong possibility of a race condition.

3. Even if you could fulfil the requires of 2 you'd have to "busy wait" (ie. keep spinning in code until your thread could move on), where as a Mutex (being a kernal object) uses up no CPU time whilst blocking.


>> However i am still confused the fact that we need to have multiple reader, against the fact that R/W cannot be run together.
The idea is it takes a fraction of the time to add to the queue than it does to process jobs in the queue. So, if might take 10ms to add a new job and dispatch a thread but then that thread might spend 20 seconds processing the job. Meanwhile, the producing can keep adding jobs and dispatching new threads until all the threads in the pool are busy, at which point the jobs will just stake queued until a thread is free to process it. Of course, if you only want one thread in your thread pool then you can just set up the main process to only create a thread pool of 1 thread.


Author Closing Comment

ID: 31577341

Expert Comment

ID: 24379625

Featured Post

Networking for the Cloud Era

Join Microsoft and Riverbed for a discussion and demonstration of enhancements to SteelConnect:
-One-click orchestration and cloud connectivity in Azure environments
-Tight integration of SD-WAN and WAN optimization capabilities
-Scalability and resiliency equal to a data center

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
javap bin 2 40
Constant string is of type char *   ? 7 42
ejb on wildfly 5 30
windows 10 pro lost profile. 10 43
Go is an acronym of golang, is a programming language developed Google in 2007. Go is a new language that is mostly in the C family, with significant input from Pascal/Modula/Oberon family. Hence Go arisen as low-level language with fast compilation…
Basic understanding on "OO- Object Orientation" is needed for designing a logical solution to solve a problem. Basic OOAD is a prerequisite for a coder to ensure that they follow the basic design of OO. This would help developers to understand the b…
Video by: Grant
The goal of this video is to provide viewers with basic examples to understand and use nested-loops in the C programming language.
The goal of this video is to provide viewers with basic examples to understand and use conditional statements in the C programming language.

790 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question