小编Fla*_*gas的帖子

我有一个std::packaged_task包含lambda的lambda,它通过复制捕获变量.当这个std::packaged_task被删除,我希望在lambda里面的变量生活是破坏,但我注意到,如果我得到了相关std::future为此std::packaged_task,该future对象扩展拉姆达内部变量的生存期.

例如:

#include <iostream>
#include <future>

class Dummy
{
public:
    Dummy() {std::cout << this << ": default constructed;" << std::endl;}
    Dummy(const Dummy&) {std::cout << this << ": copy constructed;" << std::endl;}
    Dummy(Dummy&&) {std::cout << this << ": move constructed;" << std::endl;}
    ~Dummy() {std::cout << this << ": destructed;" << std::endl;}
};

int main()
{
    std::packaged_task<void()>* p_task;
    {
        Dummy ScopedDummy;
        p_task = new std::packaged_task<void()>([ScopedDummy](){std::cout << "lambda call with: " << &ScopedDummy …