如何将 unique_ptr 捕获到 lambda 表达式中？

Question

I have tried the following:

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
    };
}

But it does not compile. Any ideas?

UPDATE:

AS suggested, some new syntax is required to explicitly specify we need to transfer the ownership to the lambda, I am now thinking about the following syntax:

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
    };
}

Would it be a good candidate?

UPDATE 1:

I will show my implementation of move and copy as following:

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

This issue is addressed by lambda generalized capture in C++14:

// 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);});

Answer 2

You cannot permanently capture a unique_ptr in a lambda. Indeed, if you want to permanently capture anything in a lambda, it must be copyable ; merely movable is insufficient.

This could be considered a defect in C++11, but you would need some syntax to explicitly say that you wanted to move the unique_ptr value into the lambda. The C++11 specification is very carefully worded to prevent implicit moves on named variables; that's why std::move exists, and this is a good thing.

To do what you want will require either using std::bind (which would be semi-convoluted, requiring a short sequence of binds ) or just returning a regular old object.

Also, never take unique_ptr by && , unless you are actually writing its move constructor. Just take it by value; the only way a user can provide it by value is with a std::move . Indeed, it's generally a good idea to never take anything by && , unless you're writing the move constructor/assignment operator (or implementing a forwarding function).

Answer 3

The "semi-convoluted" solution using std::bind as mentioned by Nicol Bolas is not so bad after all:

std::function<void ()> getAction(std::unique_ptr<MyClass>&& psomething)
{
    return std::bind([] (std::unique_ptr<MyClass>& p) { p->do_some_thing(); },
                     std::move(psomething));
}

Answer 4

A sub-optimal solution that worked for me was to convert the unique_ptr to a shared_ptr and then capture the shared_ptr in the lambda.

std::function<void()> getAction(std::unique_ptr<MyClass> psomething)
{
    //The caller given ownership of psomething
    std::shared_ptr<MyClass> psomethingShared = std::shared_ptr<MyClass>(std::move(psomething));
    return [psomethingShared]()
    {
        psomethingShared->do_some_thing();
    };
}

Answer 5

I used this really dodgy workaround, which involves sticking the unique_ptr inside a shared_ptr . This is because my code required a unique_ptr (due to an API restriction) so I couldn't actually convert it to a shared_ptr (otherwise I'd never be able to get my unique_ptr back).

My justification for using this abomination is that it was for my test code, and I had to std::bind a unique_ptr into the test function call.

// Put unique_ptr inside a shared_ptr
auto sh = std::make_shared<std::unique_ptr<Type>>(std::move(unique));

std::function<void()> fnTest = std::bind([this, sh, input, output]() {
    // Move unique_ptr back out of shared_ptr
    auto unique = std::move(*sh.get());

    // Make sure unique_ptr is still valid
    assert(unique);

    // Move unique_ptr over to final function while calling it
    this->run_test(std::move(unique), input, output);
});

Now calling fnTest() will call run_test() while passing the unique_ptr to it. Calling fnTest() a second time will result in an assertion failure, because the unique_ptr has already been moved/lost during the first call.

Answer 6

One also need to know, that lambdas capturing unique_ptr cannot be converted into std::function because std::function requires that the callable object is copyable.

auto lambdaWithoutCapture = [](){return 1;}; //Can be std::function
auto lambdaWithCapture = [=](){return 1;}; //Can be std::function
auto lambdaWithCapture2 = [&](){return 1;}; //Can be std::function
auto lambdaWithCapture3 = [uptrProblematic = std::move(uptrProblematic)]() mutable {return 1;}; //Can't be std::function

Therefore, if you don't have to specify return type of the function, you can use such approach which does not use std::function. But you need to know, that this will only work in local scope. You can't declare auto workerFactory(); in header file, as this will raise compilation error.

auto workerFactory()
{
    std::unique_ptr uptrProblematic = std::make_unique<int>(9);
    int importantData = 71;
    return [=, uptrProblematic = std::move(uptrProblematic)](std::string input) mutable -> int {
        std::cout << "Problematic variable is equal to: " << *uptrProblematic << "\n";
        std::cout << "Important data is equal to: " << importantData << "\n";
        std::cout << "Input equal to: " << input << "\n";
        return 9;
    };
}

int main()
{
    auto worker = workerFactory();
    worker("Test");
}