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考慮這個簡單的可變參數模板函數,該函數產生一個線程並將該參數轉發給線程函數。 爲什麼我在這裏得到線程構造函數的模板替換失敗?通過完美轉發將variadic模板參數作爲std :: thread的參考
std::thread t;
void test3(int& a)
{
a = 10;
}
template<class ...Args>
void test(Args&&... args)
{
t = std::thread(test3, std::forward<Args>(args)...);
}
int main()
{
auto timer = 2s;
int a = 1;
test(a);
std::this_thread::sleep_for(timer);
std::cout << a << std::endl;
t.join();
}
編譯器輸出:
template argument deduction/substitution failed:
/opt/wandbox/gcc-head/include/c++/8.0.0/bits/invoke.h: In substitution of
'template<class _Callable, class ... _Args> constexpr typename
std::__invoke_result<_Functor, _ArgTypes>::type std::__invoke(_Callable&&,
_Args&& ...) [with _Callable = void (*)(int&); _Args = {int}]':
/opt/wandbox/gcc-head/include/c++/8.0.0/thread:233:29: required by
substitution of 'template<long unsigned int ..._Ind> decltype
(std::__invoke(_S_declval<_Ind>()...)) std::thread::_Invoker<std::tuple<void
(*)(int&), int> >::_M_invoke<_Ind ...>(std::_Index_tuple<_Ind1 ...>) [with
long unsigned int ..._Ind = {0, 1}]'
/opt/wandbox/gcc-head/include/c++/8.0.0/thread:240:2: required from
'struct std::thread::_Invoker<std::tuple<void (*)(int&), int> >'
/opt/wandbox/gcc-head/include/c++/8.0.0/thread:127:22: required from
'std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)
(int&); _Args = {int&}]'
prog.cc:23:14: required from 'void test(Args&& ...) [with Args = {int&}]'
prog.cc:43:11: required from here
/opt/wandbox/gcc-head/include/c++/8.0.0/bits/invoke.h:89:5: error: no type
named 'type' in 'struct std::__invoke_result<void (*)(int&), int>'
當我換的,像這樣一個std :: REF參數轉發:
std::thread(test3, std::ref(std::forward<Args>(args)...));
它的工作原理。這些論據是否應該首先完美轉發?
['線程的構造函數](http://en.cppreference.com/w/cpp/thread/thread/thread)是根據decay_copy定義的。 – ildjarn
爲了詳細說明ildjarn暗示的是什麼,你傳遞的函數被調用一個引用的整數的副本。並且該副本是即將到期的值。它不能綁定到一個非const的左值引用。 – StoryTeller
@StoryTeller是唯一的解決方案,然後在參數上使用std :: ref? – Mozbi