CODEWITHSUNDEEP
c++c++11shared-ptrauto
Olumide
Olumide

Reputation: 5809

Returning a shared_ptr from a function

I'm very new to C++11, 'still very much experimenting with the extensions. I find the auto keyword very convenient, particularly when dealing with template variables. This means that given

template<typename ... Types>
struct Foo
{
};

template<typename ... Types>
Foo<Types ...>* create( Types ... types ... )
{
    return new Foo<Types ...>;
}

I can now make the assignment

auto t1 = create( 'a' , 42 , true , 1.234 , "str" );

instead of

Foo<char, int, bool, double , const char*>* t2 = create( 'a' , 42 , true , 1.234 , "str" );

The problem now is that because t1 is a pointer I'd like to hold it in a shared_ptr as Herb Sutter recommended. Therefore, I'd like to store the return value of create() in a shared_ptr without having to name the template argument types, as in t2.

Upvotes: 2

Views: 1619

Answers (2)

Olumide
Olumide

Reputation: 5809

This is too long to post as a comment. Therefore I'm posting it here instead. Besides it might be an answer.

@nosid why not the following. Its less convoluted. 

template<typename ... Types>
struct Foo
{
    Foo( Types ... types ... ) : m_data( types ...)
    {
    }

    std::tuple<Types...>    m_data;
};

template<typename ... Types>
std::shared_ptr<Foo<Types ...> > create( Types ... types ... )
{
    return std::make_shared<Foo<Types ...> >( types ... );
}

Upvotes: 0

nosid
nosid

Reputation: 50034

Avoid using raw pointers all together. Use std::make_shared and make_unique (not right in the standard) instead of new. Then auto will work nicely. E.g.

template <typename ...Args>
auto create(Args&&... args)
    -> std::shared_ptr<Foo<typename std::decay<Args>::type...>>
{
    return std::make_shared<Foo<typename std::decay<Args>::type...>>(
        std::forward<Args>(args)...);
}

Upvotes: 2

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