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互斥体可以确保对象的线程可见性,而不是明确地保护它们吗?

考虑以下代码,线程是否可能以不同的方式看到对象的状态,尽管它们都是由相同的指针引用的?

using namespace std;

class ProducerAndConsumer{

  class DummyObject {
  public:
    DummyObject() {
      sprintf(a, "%d", rand());
    }
  private:
    char a[1000];
  };

  mutex queue_mutex_;
  queue<DummyObject *> queue_;
  thread *t1, *t2;

  void Produce() {
    while (true) {
      Sleep(1);
      // constructing object without any explicit synchronization
      DummyObject *dummy = new DummyObject();
      {
        lock_guard<mutex> guard(queue_mutex_);
        if (queue_.size() > 1000) {
          delete dummy;
          continue;
        }
        queue_.push(dummy);
      }
    }
  }

  void Consume() {
    while (true) {
      Sleep(1);
      DummyObject *dummy;
      {
        lock_guard<mutex> guard(queue_mutex_);
        if (queue_.empty())
          continue;
        dummy = …
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c++ multithreading

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