How to easily make std::cout thread-safe?

Question

I have a multi-threaded application, which heavily uses std::cout for logging without any locking. In such a case, how can I easily add lock mechanism to make std::cout thread-safe?

I don't want to search for each occurrence of std::cout and add a line of locking code. That is too tedious.

Any better practice?

Answer 1

In addition to synchronisation this solution provides information about the thread from which the log was written.

DISCLAIMER: It's quite a naive way of syncing the logs, however it might be applicable for some small use cases for debugging.

thread_local int thread_id = -1;
std::atomic<int> thread_count;

struct CurrentThread {

  static void init() {
    if (thread_id == -1) {
      thread_id = thread_count++;
    }
  }

  friend std::ostream &operator<<(std::ostream &os, const CurrentThread &t) {
    os << "[Thread-" << thread_id << "] - ";
    return os;
  }
};

CurrentThread current_thread;
std::mutex io_lock;
#ifdef DEBUG
#define LOG(x) {CurrentThread::init(); std::unique_lock<std::mutex> lk(io_lock); cout << current_thread; x << endl;}
#else
#define LOG(x)
#endif

This can be used like this.

LOG(cout << "Waiting for some event");

And it will give log output

[Thread-1] - Entering critical section 
[Thread-2] - Waiting on mutex
[Thread-1] - Leaving critical section, unlocking the mutex

Answer 2

I guess you could implement your own class which wraps cout and associates a mutex with it. The operator << of that new class would do three things:

1. create a lock for the mutex, possibly blocking other threads
2. do the output, i.e. do the operator << for the wrapped stream and the passed argument
3. construct an instance of a different class, passing the lock to that

This different class would keep the lock and delegate operator << to the wrapped stream. The destructor of that second class would eventually destroy the lock and release the mutex.

So any output you write as a single statement, i.e. as a single sequence of << invocations, will be printed atomically as long as all your output goes through that object with the same mutex.

Let's call the two classes synchronized_ostream and locked_ostream. If sync_cout is an instance of synchronized_ostream which wraps std::cout, then the sequence

sync_cout << "Hello, " << name << "!" << std::endl;

would result in the following actions:

1. synchronized_ostream::operator<< would aquire the lock
2. synchronized_ostream::operator<< would delegate the printing of "Hello, " to cout
3. operator<<(std::ostream&, const char*) would print "Hello, "
4. synchronized_ostream::operator<< would construct a locked_ostream and pass the lock to that
5. locked_ostream::operator<< would delegate the printing of name to cout
6. operator<<(std::ostream&, std::string) would print the name
7. The same delegation to cout happens for the exclamation point and the endline manipulator
8. The locked_ostream temporary gets destructed, the lock is released

Answer 3

Along the lines of the answer suggested by Conchylicultor, but without inheriting from std::ostringstream:

---

EDIT: Fixed return type for the overloaded operator and added overload for std::endl.

---

EDIT 1: I have extended this into a simple header-only library for logging / debugging multi-threaded programs.

---

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

static std::mutex mtx_cout;

// Asynchronous output
struct acout
{
        std::unique_lock<std::mutex> lk;
        acout()
            :
              lk(std::unique_lock<std::mutex>(mtx_cout))
        {

        }

        template<typename T>
        acout& operator<<(const T& _t)
        {
            std::cout << _t;
            return *this;
        }

        acout& operator<<(std::ostream& (*fp)(std::ostream&))
        {
            std::cout << fp;
            return *this;
        }
};

int main(void)
{


    std::vector<std::thread> workers_cout;
    std::vector<std::thread> workers_acout;

    size_t worker(0);
    size_t threads(5);


    std::cout << "With std::cout:" << std::endl;

    for (size_t i = 0; i < threads; ++i)
    {
        workers_cout.emplace_back([&]
        {
            std::cout << "\tThis is worker " << ++worker << " in thread "
                      << std::this_thread::get_id() << std::endl;
        });
    }
    for (auto& w : workers_cout)
    {
        w.join();
    }

    worker = 0;

    std::this_thread::sleep_for(std::chrono::seconds(2));

    std::cout << "\nWith acout():" << std::endl;

    for (size_t i = 0; i < threads; ++i)
    {
        workers_acout.emplace_back([&]
        {
            acout() << "\tThis is worker " << ++worker << " in thread "
                    << std::this_thread::get_id() << std::endl;
        });
    }
    for (auto& w : workers_acout)
    {
        w.join();
    }

    return 0;
}

Output:

With std::cout:
        This is worker 1 in thread 139911511856896
        This is worker  This is worker 3 in thread 139911495071488
        This is worker 4 in thread 139911486678784
2 in thread     This is worker 5 in thread 139911503464192139911478286080


With acout():
        This is worker 1 in thread 139911478286080
        This is worker 2 in thread 139911486678784
        This is worker 3 in thread 139911495071488
        This is worker 4 in thread 139911503464192
        This is worker 5 in thread 139911511856896

Answer 4

Since C++20, you can use std::osyncstream wrapper:

http://en.cppreference.com/w/cpp/io/basic_osyncstream

{
  std::osyncstream bout(std::cout); // synchronized wrapper for std::cout
  bout << "Hello, ";
  bout << "World!";
  bout << std::endl; // flush is noted, but not yet performed
  bout << "and more!\n";
} // characters are transferred and std::cout is flushed

It provides the guarantee that all output made to the same final destination buffer (std::cout in the examples above) will be free of data races and will not be interleaved or garbled in any way, as long as every write to the that final destination buffer is made through (possibly different) instances of std::basic_osyncstream.

Answer 5

I know its an old question, but it helped me a lot with my problem. I created an utility class based on this post answers and I'd like to share my result.

Considering we use C++11 or latter C++ versions, this class provides print and println functions to compose strings before calling the standard output stream and avoid concurrency problems. These are variadic functions which use templates to print different data types.

You can check its use in a producer-consumer problem on my github: https://github.com/eloiluiz/threadsBar

So, here is my code:

class Console {
private:
    Console() = default;

    inline static void innerPrint(std::ostream &stream) {}

    template<typename Head, typename... Tail>
    inline static void innerPrint(std::ostream &stream, Head const head, Tail const ...tail) {
        stream << head;
        innerPrint(stream, tail...);
    }

public:
    template<typename Head, typename... Tail>
    inline static void print(Head const head, Tail const ...tail) {
        // Create a stream buffer
        std::stringbuf buffer;
        std::ostream stream(&buffer);
        // Feed input parameters to the stream object
        innerPrint(stream, head, tail...);
        // Print into console and flush
        std::cout << buffer.str();
    }

    template<typename Head, typename... Tail>
    inline static void println(Head const head, Tail const ...tail) {
        print(head, tail..., "\n");
    }
};

Answer 6

For fast debugging c++11 applications and avoid interleaved output I just write small functions like these:

...
#include <mutex>
...
mutex m_screen;
...
void msg(char const * const message);
...
void msg(char const * const message)
{
  m_screen.lock();
  cout << message << endl;
  m_screen.unlock();
}

I use these types of functions for outputs and if numeric values are needed I just use something like this:

void msgInt(char const * const message, int const &value);
...
void msgInt(char const * const message, int const &value)
{
  m_screen.lock();
  cout << message << " = " << value << endl;
  m_screen.unlock();
}

This is easy and works fine to me, but I don't really know if it is technically correct. So I would be glad to hear your opinions.

---

Well, I didn't read this:

I don't want to search for each occurrence of std::cout and add a line of locking code.

I'm sorry. However I hope it helps somebody.