我与你分享了一个问题,我用类使用可变参数函数得到了一个问题。它是以下代码中显示的类Thread。它是std :: thread的一个包装,以便使用函数模式。C++多态性与可变参数函数
我想在继承Thread类进入一个新的类,仿函数使用多态性与这个功能,但是GCC返回错误吼叫:
#include <thread>
#include <iostream>
using namespace std;
template<class... Args>
class Thread
{
public:
virtual void operator()(Args...) = 0;
void run(Args... args)
{
std::thread t(std::forward< Thread<Args...> >(*this), std::forward<Args>(args)...);
t.join();
}
};
template<class... Args>
class Functor : public Thread<Args...>
{
public:
// generates the errors bellow
virtual void operator()(Args... /*args*/)
{
}
// doesnot work since the pure virtual function wants another prototype of function.
// void operator()(int)
// {
// }
};
int main()
{
int a = 12;
Functor<int> f;
f.run(ref(a));
return 0;
}
from t-Thread-args2.cpp:1: /usr/lib/gcc/x86_64-unknown-linux-gnu/4.7.2/../../../../include/c++/4.7.2/tuple: In instantiation of ‘struct std::_Head_base, false>’: /usr/lib/gcc/x86_64-unknown-linux-gnu/4.7.2/../../../../include/c++/4.7.2/tuple:215:12: required from ‘struct std::_Tuple_impl, int>’ /usr/lib/gcc/x86_64-unknown-linux-gnu/4.7.2/../../../../include/c++/4.7.2/tuple:507:11: required from ‘class std::tuple, int>’ /usr/lib/gcc/x86_64-unknown-linux-gnu/4.7.2/../../../../include/c++/4.7.2/functional:1601:39: required from ‘struct std::_Bind_simple(int)>’ /usr/lib/gcc/x86_64-unknown-linux-gnu/4.7.2/../../../../include/c++/4.7.2/thread:133:9: required from ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = Thread; _Args = {int}]’ t-Thread-args2.cpp:14:83: required from ‘void Thread::run(Args ...) [with Args = {int}]’ t-Thread-args2.cpp:42:17: required from here /usr/lib/gcc/x86_64-unknown-linux-gnu/4.7.2/../../../../include/c++/4.7.2/tuple:166:13: error: cannot declare field ‘std::_Head_base, false>::_M_head_impl’ to be of abstract type ‘Thread’ t-Thread-args2.cpp:7:7: note: because the following virtual functions are pure within ‘Thread’: t-Thread-args2.cpp:10:18: note: void Thread::operator()(Args ...) [with Args = {int}]
我真的不理解,因为纯错误虚拟函数在派生类中被很好地定义。但是,在将函数run()移入派生类(Functor)中时,它起作用。
由于提前, 蚕儿
“std :: forward”的用法没有意义。它对函数模板有意义,但是在你的代码中,run()不是函数模板。 – Nawaz 2013-02-21 07:51:01
更具体地说,'std :: forward'在这里没有意义,因为它的参数总是'* this',即总是一个左值。 – jogojapan 2013-02-21 07:57:45
@jogojapan:“std :: forward”的参数几乎总是左值,因为它通常有一个*名称*。 – Nawaz 2013-02-21 08:03:28