How to capture a unique_ptr into a lambda expression? - c++

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");
}