c++ - 静态线程函数访问 C++ 中的非静态类成员

Question

Class Test{
    int value;
    static void* thread_func(void* args){
        value++;
    }
    void newthread(){
        pthread_create(&thread_func,...);
    }
}

我试图在 Class Test. 中创建一个线程。因此编译器强制我将 thread_func 设为静态。但是，我无法再访问非静态成员“value”。它说:

invalid use of member 'Class::value' in static member function

有办法解决吗？

Answer 1

However I cannot access the non-static member "value" anymore.

那是因为你的类中的 static 函数没有(而且不能有)this 指针。您只需要将指向您的 Test 对象的指针传递给 pthread_create()作为第四个参数，然后执行此操作:

static void* thread_func(void* args)
{
      Test *test = static_cast<Test*>(args);
      test->value++;
      //write return statement properly
}

---

但是，如果您在 thread_func() 中做太多需要在许多地方访问 Test 类成员的事情，那么我会建议这种设计:

//this design simplifies the syntax to access the class members!
class Test
{
    //code omitted for brevity

    static void* thread_func(void* args)
    {
          Test *test = static_cast<Test*>(args);
          test->run(); //call the member function!
          //write return statement properly
    }
    void run() //define a member function to run the thread!
    { 
          value++;//now you can do this, because it is same as 'this->value++;
          //you do have 'this' pointer here, as usual; 
          //so access other members like 'value++'.
    }
    //code omitted for brevity
  }

---

更好的设计:定义一个可重用的类!

更好的做法是定义一个可重用类，其中包含纯虚函数run()，由派生类实现。它应该是这样设计的:

//runnable is reusable class. All thread classes must derive from it! 
class runnable
{
public:
    virtual ~runnable() {}
    static void run_thread(void *args)
    {
        runnable *prunnable = static_cast<runnable*>(args);
        prunnable->run();
    }
protected:
    virtual void run() = 0; //derived class must implement this!
};

class Test : public runnable //derived from runnable!
{
public:
    void newthread()
    {
        //note &runnable::run_thread
        pthread_create(&runnable::run_thread,..., this);
    }
protected:
    void run() //implementing the virtual function!
    {
        value++; // your thread function!
    }
}

看起来更好？