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C++ Cookbook
book

C++ Cookbook

by D. Ryan Stephens, Christopher Diggins, Jonathan Turkanis, Jeff Cogswell
November 2005
Beginner to intermediate content levelBeginner to intermediate
594 pages
16h 23m
English
O'Reilly Media, Inc.
Content preview from C++ Cookbook

12.2. Making a Resource Thread-Safe

Problem

You are using multiple threads in a program and you need to ensure a resource is not modified by more than one thread at a time. In general, this process is called making the resource thread-safe, or serializing access to it.

Solution

Use the class mutex, defined in boost/thread/mutex.hpp, to synchronize access among threads. Example 12-2 shows a simple use of a mutex object to control concurrent access to a queue.

Example 12-2. Making a class thread-safe

#include <iostream> #include <boost/thread/thread.hpp> #include <string> // A simple queue class; don't do this, use std::queue template<typename T> class Queue { public: Queue() {} ~Queue() {} void enqueue(const T& x) { // Lock the mutex for this queue boost::mutex::scoped_lock lock(mutex_); list_.push_back(x); // A scoped_lock is automatically destroyed (and thus unlocked) // when it goes out of scope } T dequeue() { boost::mutex::scoped_lock lock(mutex_); if (list_.empty()) throw "empty!"; // This leaves the current scope, so the T tmp = list_.front(); // lock is released list_.pop_front(); return(tmp); } // Again: when scope ends, mutex_ is unlocked private: std::list<T> list_; boost::mutex mutex_; }; Queue<std::string> queueOfStrings; void sendSomething() { std::string s; for (int i = 0; i < 10; ++i) { queueOfStrings.enqueue("Cyrus"); } } void recvSomething() { std::string s; for (int i = 0; i < 10; ++i) { try {s = queueOfStrings.dequeue();} catch(...) {} } } int main() { boost::thread ...
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Publisher Resources

ISBN: 0596007612Supplemental ContentErrata Page