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Home/ Questions/Q 3595868
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T19:54:33+00:00 2026-05-18T19:54:33+00:00

I have a class with a templated constructor for implicit move conversion, however this

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I have a class with a templated constructor for implicit move conversion, however this constructor should NOT be used for the class (which should only be copy constructible). However, the compiler always tries to use the templated constructor instead of the regular copy constructor.

e.g. With this i get the follow compiler errors, link. (you can just copy paste this code if you want to try it)

struct implementation{};
class my_class
{
 my_class(my_class&&); // delete move-constructor... OUCH... COMPILER ERROR
public:
 my_class(){}
 my_class(const my_class& other) : impl_(other.impl_){}

 template<typename T>
 my_class(T&& impl) : impl_(std::make_shared<T>(std::move(impl))){} // Still tries to use this...

private:
 std::shared_ptr<implementation> impl_;
};

class other_class
{
public:
 my_class foo()
 { 
       return instance_; // Wants to use move-constructor???
 }
private:
 my_class instance_;
};

Any one got an idea how to solve this properly?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-18T19:54:34+00:00Added an answer on May 18, 2026 at 7:54 pm

    Okay, here is my complete overhaul of my_class:

    class my_class
    {
    public:
        my_class() {}
        my_class(my_class&& that) : impl_(std::move(that.impl_)) {}
    
        template <typename T> my_class(T&& impl,
        typename std::enable_if<
            std::is_base_of<
                implementation,
                typename std::remove_reference<T>::type
            >::value,
            void
        >::type* dummy = 0
        ) : impl_(std::make_shared<T>(std::forward<T>(impl))) {}
    
        template <typename T>
        typename std::enable_if<
            std::is_base_of<
                implementation,
                typename std::remove_reference<T>::type
            >::value,
            my_class&
        >::type
        operator=(T&& impl)
        {
            std::make_shared<implementation>(std::forward<T>(impl)).swap(impl_);
            return *this;
        }
    
    private:
        std::shared_ptr<implementation> impl_;
    };
    

    As suggested by others, this works for lvalue and rvalue references by using std::forward instead of std::move. The remove_reference is necessary because for lvalue references, T is a reference, and derived& does not derive from base, but derived does (note the reference).

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