C++ 使用没有动态分配的接口(interface)

Question

我正在尝试使用接口(interface)

class IDemo
{
    public:
        virtual ~IDemo() {}
        virtual uint32_t Some_Operation(uint32_t a, uint32_t b) = 0;
};





class Child_A : public IDemo
{
    public:
        virtual uint32_t Some_Operation(uint32_t a, uint32_t b);
};



class Child_B : public IDemo
{
    public:
        virtual uint32_t Some_Operation(uint32_t a, uint32_t b);
};

Child_A::Some_Operation 返回 a+b 的总和 Child_B::一些操作返回产品a*b

用法如下

bool Test_Inferface()
{

    IDemo* pDemo = new Child_B();

    uint32_t product = pDemo->Some_Operation(1, 2);

    delete pDemo;

    if (2 != product)
    {
        return false;
    }


    pDemo = new Child_A();

    uint32_t sum = pDemo->Some_Operation(1,2);

    delete pDemo;

    if(3 != sum)
    {
        return false;
    }

    return true;
}

由于可能的内存泄漏，我试图避免新建/删除。 是否可以静态分配接口(interface)？

IDemo test = Child_A();

编译器不喜欢那样。

Answer 1

非常简单。多态性(接口(interface))与动态分配无关。如果你不想动态分配，那就不要。

您的示例很容易像这样工作:

bool Test_Inferface()
{

    Child_B child_b;

    uint32_t product = child_b.Some_Operation(1, 2);

    if (2 != product)
    {
        return false;
    }


    Child_A chile_a;

    uint32_t sum = child_a.Some_Operation(1,2);

    if(3 != sum)
    {
        return false;
    }

    return true;
}

您可能需要一个实际使用接口(interface)的更好示例。

// here we have an interface reference parameter that doesn't care if what is
// passed to it is dynamically allocated or sitting on the stack.
uint32_t better_example(IDemo& demo)
{
    return demo.Some_Operation(1, 2);
}

bool Test_Inferface()
{

    Child_B child_b;

    uint32_t product = better_example(child_b);

    if (2 != product)
    {
        return false;
    }


    Child_A chile_a;

    uint32_t sum = better_example(child_a);

    if(3 != sum)
    {
        return false;
    }

    return true;
}