c++ - 将 unique_ptr 返回到多态类型

Question

我试图通过从我的一些方法返回 unique_ptr 而不是原始指针来变得更安全。但是，在返回指向多态类型的唯一指针时，我有点困惑。

我们如何返回指向派生类类型的基类类型的唯一指针？

另外，作为一个不太重要的次要问题 - 我是否使用移动构造函数正确地从基类创建派生类？

这是我的最小示例:

// Standard Includes
#include <exception>
#include <memory>
#include <string>
#include <sstream>

//--------------------------------------------------------------------------------------------------
class BaseResult
{
public:

    std::string m_x;

    virtual ~BaseResult() {};
};

class DerivedResult : public BaseResult
{
public:

    int m_y;

    DerivedResult()
        :
        BaseResult()
    {}

    DerivedResult(const DerivedResult & rhs)
        :
        BaseResult(rhs)
      , m_y       (rhs.m_y)
    {}

    DerivedResult(DerivedResult && rhs)
        :
        BaseResult(std::move(rhs))
      , m_y(rhs.m_y)
    {}

    DerivedResult(BaseResult && rhs)
        :
        BaseResult(std::move(rhs))
        , m_y()
    {
    }

    ~DerivedResult() {}
};

class BaseCalc
{
public:

    virtual ~BaseCalc() {}

    virtual std::unique_ptr<BaseResult> Calc() const
    {
        std::unique_ptr<BaseResult> result(new BaseResult);
        result->m_x = "Base Calced";

        return result;
    }

};

class DerivedCalc : public BaseCalc
{
public:

    virtual ~DerivedCalc() {}

    virtual std::unique_ptr<BaseResult> Calc() const
    {
        // I need to rely on the base calculations to get the fields relevant to the base
        std::unique_ptr<BaseResult> baseResult = BaseCalc::Calc();

        // However, I want to perform my addtional calculation relevant to derived, here.
        std::unique_ptr<DerivedResult> result(new DerivedResult(std::move(*baseResult)));
        result->m_y = 2;

        /* Results in error
        return result;
        */
        return std::move(result); // Is this how we do it?
    }
};

//--------------------------------------------------------------------------------------------------
int main()
{
    DerivedCalc calculator;
    std::unique_ptr<BaseResult> temp = calculator.Calc();

    // Cast - Got this part from https://stackoverflow.com/questions/21174593/
    std::unique_ptr<DerivedResult> actualResult;

    if (DerivedResult * cast = dynamic_cast<DerivedResult *>(temp.get()))
    {
        actualResult = std::unique_ptr<DerivedResult>(cast, std::move(temp.get_deleter()));
        temp.release();
    }
    else
    {
        std::exception("Failed to cast to DerivedResult");
    }

    std::string x = actualResult->m_x;
    int y = actualResult->m_y;

    return 0;
}

Answer 1

回答您的第一个问题，“我们如何返回指向派生类类型的基类类型的唯一指针？” (注意 std::make_unique 需要 c++14):

class DerivedCalc : public BaseCalc
{
public:

    virtual ~DerivedCalc() {}

    virtual std::unique_ptr<BaseResult> Calc() const
    {
        // I need to rely on the base calculations to get the fields relevant to the base
        std::unique_ptr<BaseResult> baseResult = BaseCalc::Calc();

        // However, I want to perform my addtional calculation relevant to derived, here.
        std::unique_ptr<BaseResult> result(std::make_unique<DerivedResult>(std::move(*baseResult)));
        DerivedResult& derived = static_cast<DerivedResult&>(*result);
        derived.m_y = 2;

        return result;
    }
};

对于您的第二个问题，“我是否使用移动构造函数正确地从基类创建派生类？”，您的版本有效，但您也可以移动派生数据字段以获得最佳性能:

class DerivedResult : public BaseResult
{
    ...
    DerivedResult(DerivedResult && rhs)
        :
        BaseResult(std::move(rhs))
        , m_y(std::move(rhs.m_y))
    {}
    ...
};