c++ - 如何将 unique_ptr 捕获到 lambda 表达式中？

Question

我尝试了以下方法:

std::function<void ()> getAction(std::unique_ptr<MyClass> &&psomething){
    //The caller given ownership of psomething
    return [psomething](){ 
        psomething->do_some_thing();
        //psomething is expected to be released after this point
    };
}

但它不能编译。有什么想法吗？

更新:

正如建议的那样，需要一些新的语法来明确指定我们需要将所有权转移给 lambda，我现在正在考虑以下语法:

std::function<void ()> getAction(std::unique_ptr<MyClass> psomething){
    //The caller given ownership of psomething
    return [auto psomething=move(psomething)](){ 
        psomething->do_some_thing();
        //psomething is expected to be released after this point
    };
}

它会是一个好的候选人吗？

更新 1:

我将展示我对 move 和 copy 的实现，如下所示:

template<typename T>
T copy(const T &t) {
    return t;
}

//process lvalue references
template<typename T>
T move(T &t) {
    return std::move(t);
}

class A{/*...*/};

void test(A &&a);

int main(int, char **){
    A a;
    test(copy(a));    //OK, copied
    test(move(a));    //OK, moved
    test(A());        //OK, temporary object
    test(copy(A()));  //OK, copying temporary object
    //You can disable this behavior by letting copy accepts T &  
    //test(move(A())); You should never move a temporary object
    //It is not good to have a rvalue version of move.
    //test(a); forbidden, you have to say weather you want to copy or move
    //from a lvalue reference.
}

Answer 1

lambda generalized capture 解决了这个问题在 C++14 中:

// a unique_ptr is move-only
auto u = make_unique<some_type>(some, parameters); 

// move the unique_ptr into the lambda
go.run([u = move(u)]{do_something_with(u);});