为C ++ STL线程正确使用functor [英] correct use of functor for C++ STL thread
问题描述
我在理解功能对象作为C ++ STL中的线程例程的正确用法时遇到了一些困难.根据我的理解,函子的好处之一是对象实例可以维护状态.有时候,我希望一个或多个线程运行一些例程并计算一些结果.然后,我在加入线程后从对象查询那些结果.我正在尝试对C ++ STL线程执行相同的操作,并遇到一些问题.看来问题出在C ++ STL线程复制了我的对象,因此我不确定加入线程时应该如何检查结果.这是代码段:
I'm having some difficulty understanding the correct usage of a functional object as a thread routine in C++ STL. From my understanding, one of the benefits of a functor is that the object instance can maintain state. There are times when I want one or more threads to run some routine and compute some result. I then query those results from the objects after I have joined the threads. I'm trying to do the same with C++ STL threads and running into some problems. It appears the problem stems from the fact that the C++ STL thread makes a copy of my object and thus I'm not sure how I'm supposed to go about inspecting the results when I join the threads. Here's a snippet of the code:
#include <iostream>
#include <thread>
using namespace std;
class Worker
{
public:
Worker() : _value(0)
{
}
void operator()(unsigned int value);
unsigned int get_value() {return this->_value;}
private:
unsigned int _value;
};
void Worker::operator()(unsigned int value)
{
this->_value = value;
}
int main()
{
Worker worker;
thread thread(worker, 13);
thread.join();
unsigned int value = worker.get_value();
cout << "value: " << value << endl;
}
上面的例子只是我遇到的问题的简单再现.我希望worker.get_value()返回13但它返回零.我该如何实例化具有状态的对象,让线程在该对象中运行例程,然后在线程完成后查询该对象的状态?
The above example is just a simple repro of the problem I'm running into. I would expect worker.get_value() to return 13 yet it's returning zero. How do I go about instantiating an object with state, having a thread run a routine in that object, and then query the state of that object after the thread has completed?
谢谢, 尼克
推荐答案
当您按值传递时,将进行复制.因此,您可以通过引用包装传递引用:
When you pass by value you make a copy. So you can pass by reference through reference wrapper:
thread thread(std::ref(worker), 13);
或通过指针传递:
thread thread(&worker, 13);
在两种情况下,您都必须确保对象生存期足够长.
in both cases you have to make sure that object lifetime is long enough.
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