问题描述
一个简单的类
struct Test{
void display(int x)
{
printf("%d\n",x);
}
};
c++11
template<class F,class... Args>
auto enqueue(F&& f,Args&&... args) -> std::future<decltype(f(args...))>{
auto task = std::make_shared<std::packaged_task<decltype(f(args...)())>>(
std::bind(std::forward<F>(f),std::forward<Args>(args)...)
);
return task->get_future();
}
Test test;
cls->enqueue(&Test::display,&test,0);
编译器错误
candidate template ignored: substitution failure
[with F = void (TestMem::*)(int),Rest = <TestMem *,int>]: called object type 'void (TestMem::*)(int)' is not a function or function pointer
auto enqueue(F && f,Rest&&... rest) ->std::future<decltype(f(rest...))> {
以上在非类成员函数上运行良好,有人可以在 c++11 上提供修复
c++14 工作正常,但我的项目需要 c++11
template<class F,Args&&... args) {
using return_type = typename std::result_of<F(Args...)>::type;
auto task = std::make_shared<std::packaged_task<return_type()>>(
std::bind(std::forward<F>(f),std::forward<Args>(args)...)
);
return task->get_future();
}
解决方法
将返回类型定义为 std::future