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Home/ Questions/Q 9307575
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Editorial Team
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Editorial Team
Asked: June 19, 20262026-06-19T00:25:26+00:00 2026-06-19T00:25:26+00:00

class T {}; class UseT { public: //… boost::shared_ptr<const T> getT() const { return

  • 0
class T
{};

class UseT
{
public:
    //...
    boost::shared_ptr<const T> getT() const
    {
        return m_t;
    }
private:
    boost::shared_ptr<T> m_t;
};

Question> What are the rules used when we convert from boost::shared_ptr<T> to boost::shared_ptr<const T>?

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  1. Editorial Team
    Editorial Team
    2026-06-19T00:25:26+00:00Added an answer on June 19, 2026 at 12:25 am

    shared_ptr<T> has a converting constructor that allows it to be constructed from shared_ptr<U> if it would be valid to convert from U* to T*, mirroring how built-in pointers work.

    template<typename U>
      shared_ptr(const shared_ptr<U>& other);
    

    (For std::shared_ptr the constructor can only be called if U* is convertible to T*, but for boost::shared_ptr I’m not sure if it checks that, or you just get a compiler error for invalid conversions.)

    Since T* can be converted to const T*, the constructor allows you to create a shared_ptr<const T> from a shared_ptr<T>.

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