我正在构建一个实时软件，我在main()上有一个主要的无限循环，用于读取和处理数据的线程。
其中一个问题是保持std::vector个正在运行的线程向它们发送信号并监视执行情况。所以我把这段代码放在一起：
#include <iostream>
#include <string>
#include <vector>
#include <thread>
#include <chrono>
namespace readerThread {
void start(int id)
{
while (1)
{
std::cout << "Reader " << id << " running..." << std::endl;
std::this_thread::sleep_for(std::chrono::milliseconds(1000));
}
}
}
int main()
{
int readers[] = { 1, 2, 3 };
std::vector<std::thread> readerThreads;
for (int &reader : readers)
{
std::thread th(readerThread::start, reader);
readerThreads.push_back(th);
}
while(true)
{
std::cout << "Waiting..." << std::endl;
std::this_thread::sleep_for(std::chrono::milliseconds(10000));
}
return 0;
}
它甚至没有编译，得到这个错误：
In file included from /usr/local/include/c++/5.1.0/x86_64-unknown-linux-gnu/bits/c++allocator.h:33:0,
from /usr/local/include/c++/5.1.0/bits/allocator.h:46,
from /usr/local/include/c++/5.1.0/string:41,
from /usr/local/include/c++/5.1.0/bits/locale_classes.h:40,
from /usr/local/include/c++/5.1.0/bits/ios_base.h:41,
from /usr/local/include/c++/5.1.0/ios:42,
from /usr/local/include/c++/5.1.0/ostream:38,
from /usr/local/include/c++/5.1.0/iostream:39,
from main.cpp:1:
/usr/local/include/c++/5.1.0/ext/new_allocator.h: In instantiation of 'void __gnu_cxx::new_allocator<_Tp>::construct(_Up*, _Args&& ...) [with _Up = std::thread; _Args = {const std::thread&}; _Tp = std::thread]':
/usr/local/include/c++/5.1.0/bits/alloc_traits.h:256:4: required from 'static std::_Require<std::allocator_traits<_Alloc>::__has_construct<_Tp, _Args ...> > std::allocator_traits<_Alloc>::_S_construct(_Alloc&, _Tp*, _Args&& ...) [with _Tp = std::thread; _Args = {const std::thread&}; _Alloc = std::allocator<std::thread>; std::_Require<std::allocator_traits<_Alloc>::__has_construct<_Tp, _Args ...> > = void]'
/usr/local/include/c++/5.1.0/bits/alloc_traits.h:402:16: required from 'static decltype (_S_construct(__a, __p, (forward<_Args>)(std::allocator_traits::construct::__args)...)) std::allocator_traits<_Alloc>::construct(_Alloc&, _Tp*, _Args&& ...) [with _Tp = std::thread; _Args = {const std::thread&}; _Alloc = std::allocator<std::thread>; decltype (_S_construct(__a, __p, (forward<_Args>)(std::allocator_traits::construct::__args)...)) = <type error>]'
/usr/local/include/c++/5.1.0/bits/stl_vector.h:917:30: required from 'void std::vector<_Tp, _Alloc>::push_back(const value_type&) [with _Tp = std::thread; _Alloc = std::allocator<std::thread>; std::vector<_Tp, _Alloc>::value_type = std::thread]'
main.cpp:37:30: required from here
/usr/local/include/c++/5.1.0/ext/new_allocator.h:120:4: error: use of deleted function 'std::thread::thread(const std::thread&)'
{ ::new((void *)__p) _Up(std::forward<_Args>(__args)...); }
^
In file included from main.cpp:4:0:
/usr/local/include/c++/5.1.0/thread:126:5: note: declared here
thread(const thread&) = delete;
^
线程是独立的，所以我不需要在主程序上也不需要在任何线程上调用join ...
所以，这是我的怀疑：
为什么我的代码无法编译？
这是存储线程向量的正确方法吗？
感谢您的帮助...
答案 0 :(得分：11)
您需要使用类似
的内容readerThreads.push_back(move(th));
这将使th成为右值，并导致调用移动ctor。 thread的副本是disabled设计的（参见Anthony Williams'C++ Concurrency In Action）。
答案 1 :(得分：6)
/usr/local/include/c++/5.1.0/ext/new_allocator.h: In instantiation of 'void __gnu_cxx::new_allocator<_Tp>::construct(_Up*, _Args&& ...) [with _Up = std::thread; _Args = {const std::thread&}; _Tp = std::thread]':
/usr/local/include/c++/5.1.0/bits/alloc_traits.h:256:4: required from 'static std::_Require<std::allocator_traits<_Alloc>::__has_construct<_Tp, _Args ...> > std::allocator_traits<_Alloc>::_S_construct(_Alloc&, _Tp*, _Args&& ...) [with _Tp = std::thread; _Args = {const std::thread&}; _Alloc = std::allocator<std::thread>; std::_Require<std::allocator_traits<_Alloc>::__has_construct<_Tp, _Args ...> > = void]'
/usr/local/include/c++/5.1.0/bits/alloc_traits.h:402:16: required from 'static decltype (_S_construct(__a, __p, (forward<_Args>)(std::allocator_traits::construct::__args)...)) std::allocator_traits<_Alloc>::construct(_Alloc&, _Tp*, _Args&& ...) [with _Tp = std::thread; _Args = {const std::thread&}; _Alloc = std::allocator<std::thread>; decltype (_S_construct(__a, __p, (forward<_Args>)(std::allocator_traits::construct::__args)...)) = <type error>]'
/usr/local/include/c++/5.1.0/bits/stl_vector.h:917:30: required from 'void std::vector<_Tp, _Alloc>::push_back(const value_type&) [with _Tp = std::thread; _Alloc = std::allocator<std::thread>; std::vector<_Tp, _Alloc>::value_type = std::thread]'
main.cpp:37:30: required from here
/usr/local/include/c++/5.1.0/ext/new_allocator.h:120:4: error: use of deleted function 'std::thread::thread(const std::thread&)'
{ ::new((void *)__p) _Up(std::forward<_Args>(__args)...); }
让我们把它剥掉一点。
error: use of deleted function 'std::thread::thread(const std::thread&)'
您的代码正在尝试引入std::thread。
required from 'void std::vector<_Tp, _Alloc>::push_back(const value_type&)
push_back是罪魁祸首。
std::thread不可复制 - 复制线程意味着什么？
std::thread t1([](){});
std::thread t2 = t1;
因此std::thread个对象的实例旨在成为唯一所有者。除了简单的混乱之外，还会产生很多痛苦。
std::thread t1([](){});
std::thread t2 = std::move(t1);
t1不再是有效的线程描述符，它描述的线程现在由t2拥有。
要将此类内容放入容器，您可以使用std::move或std::emplace / std::emplace_back。
std::vector<std::thread> threads;
threads.push_back(std::move(std::thread([](){})));
threads.emplace_back([](){});
虽然您的代码专注于此特定问题，但我要指出，C ++标准将其声明为在线程仍然连接且未连接时调用线程析构函数的错误。
int main() {
std::thread t1([](){ while (true) { std::this_thread::yield(); } };
}
当main终止时，调用t1 .~thread（），它检测到线程仍然连接但未连接，这会引发异常，导致关机崩溃。
您需要join()线程，等待它终止运行，或detach()它。如果你想使用join()，你需要一些方法告诉线程停止，如果你detach()程序可能会在线程的中间退出，比如写数据等，你可能会引入一个严重的错误。
#include <thread>
#include <chrono>
#include <future>
int main () {
std::promise<void> cnx_promise;
std::shared_future<void> cnx_future;
std::thread t1([cnx_future]() {
while (cnx_future.valid()) {
std::this_thread::yield();
}
});
std::this_thread::sleep_for(std::chrono::seconds(1));
cnx_promise.set_value();
t1.join();
}
这里我们使用一个承诺，让线程知道何时停止运行，但你可以使用条件变量，信号等，甚至只是一个你测试为false的简单std::atomic<bool> ok_to_run { true };
答案 2 :(得分：2)
另一个有效的变体是在vector.push_back调用中创建线程对象。在这种情况下无需调用std :: move，因为它已经是一个右值（因此它将被移动）。
for (int &reader : readers)
readerThreads.push_back(std::thread(readerThread::start, reader));
答案 3 :(得分：0)
这保证了将在vector的存储中创建每个线程。没有对象字节将被复制。
import pygame
pygame.init()
clock = pygame.time.Clock()
screen=pygame.display.set_mode((width,height))
while True:
msElapsed = clock.tick(100)
screen.fill((0,0,0,255))
pygame.draw.circle(screen,(30,224,33,100),(250,100),10)
pygame.display.update()
for event in pygame.event.get():
if event.type==pygame.QUIT:
exit()
在通常情况下，这与@Seth建议不同。但在当前情况下，它是99％相同。