c++11特性之std::thread–进阶二
最后更新于：2022-04-01 06:32:13
继续C++11的std::thread之旅！
下面讨论如何**给线程传递参数**
这个例子是传递一个string
~~~
#include <iostream>
#include <thread>
#include <string>
void thread_function(std::string s)
{
std::cout << "thread function ";
std::cout << "message is = " << s << std::endl;
}
int main()
{
std::string s = "Kathy Perry";
std::thread t(&thread_function, s);
std::cout << "main thread message = " << s << std::endl;
t.join();
return 0;
}
~~~
如果运行，我们可以从输出结果看出传递成功了。
良好编程习惯的人都知道 传递引用的效率更高，那么我们该如何做呢？
你也许会这样写：
~~~
void thread_function(std::string &s)
{
std::cout << "thread function ";
std::cout << "message is = " << s << std::endl;
s = "Justin Beaver";
}
~~~
为了确认是否传递了引用？我们在线程函数后更改这个参数，但是我们可以看到，输出结果并没有变化，即不是传递的引用。
事实上, 依然是按值传递而不是引用。为了达到目的，我们可以这么做：
~~~
std::thread t(&thread_function, std::ref(s));
~~~
这不是唯一的方法：
我们可以使用move():
~~~
std::thread t(&thread_function, std::move(s));
~~~
接下来呢，我们就要将一下线程的复制吧！！
以下代码编译通过不了：
~~~
#include <iostream>
#include <thread>
void thread_function()
{
std::cout << "thread function\n";
}
int main()
{
std::thread t(&thread_function);
std::cout << "main thread\n";
std::thread t2 = t;
t2.join();
return 0;
}
~~~
但是别着急，稍稍修改：
~~~
include <iostream>
#include <thread>
void thread_function()
{
std::cout << "thread function\n";
}
int main()
{
std::thread t(&thread_function);
std::cout << "main thread\n";
std::thread t2 = move(t);
t2.join();
return 0;
}
~~~
大功告成！！！
再聊一个成员函数吧 **std::thread::get_id()**;
~~~
int main()
{
std::string s = "Kathy Perry";
std::thread t(&thread_function, std::move(s));
std::cout << "main thread message = " << s << std::endl;
std::cout << "main thread id = " << std::this_thread::get_id() << std::endl;
std::cout << "child thread id = " << t.get_id() << std::endl;
t.join();
return 0;
}
Output:
thread function message is = Kathy Perry
main thread message =
main thread id = 1208
child thread id = 5224
~~~
聊一聊**std::thread::hardware_concurrency()**
获得当前多少个线程：
~~~
int main()
{
std::cout << "Number of threads = "
<< std::thread::hardware_concurrency() << std::endl;
return 0;
}
//输出：
Number of threads = 2
~~~
之前介绍过c++11的**lambda表达式**
我们可以这样使用：
~~~
int main()
{
std::thread t([]()
{
std::cout << "thread function\n";
}
);
std::cout << "main thread\n";
t.join(); // main thread waits for t to finish
return 0;
}
~~~