统一指针，值和智能指针的C ++模板

Question

My real example is quite big, so I will use a simplified one. Suppose I have a data-type for a rectangle:

struct Rectangle {
  int width;
  int height;

  int computeArea() {
    return width * height;
  }
}

And another type that consumes that type, for example:

struct TwoRectangles {
  Rectangle a;
  Rectangle b;
  int computeArea() {
    // Ignore case where they overlap for the sake of argument!
    return a.computeArea() + b.computeArea();
  }
};

Now, I don't want to put ownership constraints on users of TwoRectangles , so I would like to make it a template:

template<typename T>
struct TwoRectangles {
  T a;
  T b;
  int computeArea() {
    // Ignore case where they overlap for the sake of argument! 
    return a.computeArea() + b.computeArea();
  }
};

Usages:

TwoRectangles<Rectangle> x;
TwoRectangles<Rectangle*> y;
TwoRectangles<std::shared_ptr<Rectangle>> z;
// etc... 

The problem is that if the caller wants to use pointers, the body of the function should be different:

template<typename T>
struct TwoRectangles {
  T a;
  T b;
  int computeArea() {
    assert(a && b);
    return a->computeArea() + b->computeArea();
  }
};

What is the best way of unifying my templated function so that the maxiumum amount of code is reused for pointers, values and smart pointers?

Answer 1

One way of doing this, encapsulating everything within TwoRectangles , would be something like:

template<typename T>
struct TwoRectangles {
  T a;
  T b;

  int computeArea() {
    return areaOf(a) + areaOf(b);
  }

private:
    template <class U>
    auto areaOf(U& v) -> decltype(v->computeArea()) {
        return v->computeArea();
    }

    template <class U>
    auto areaOf(U& v) -> decltype(v.computeArea()) {
        return v.computeArea();
    }
};

It's unlikely you'll have a type for which both of those expressions are valid. But you can always add additional disambiguation with a second argument to areaOf() .

---

Another way, would be to take advantage of the fact that there already is a way in the standard library of invoking a function on whatever: std::invoke() . You just need to know the underlying type:

template <class T, class = void>
struct element_type {
    using type = T;
};

template <class T>
struct element_type<T, void_t<typename std::pointer_traits<T>::element_type>> {
    using type = typename std::pointer_traits<T>::element_type;
};

template <class T>
using element_type_t = typename element_type<T>::type;

and

template<typename T>
struct TwoRectangles {
  T a;
  T b;

  int computeArea() {
    using U = element_type_t<T>;
    return std::invoke(&U::computeArea, a) + 
        std::invoke(&U::computeArea, b);
  }
};

Answer 2

I actually had a similar problem some time ago, eventually i opted not to do it for now (because it's a big change), but it spawned a solution that seems to be correct.

I thought about making a helper function to access underlying value if there is any indirection. In code it would look like this, also with an example similar to yours.

#include <iostream>
#include <string>
#include <memory>

namespace detail
{
    //for some reason the call for int* is ambiguous in newer standard (C++14?) when the function takes no parameters. That's a dirty workaround but it works...
    template <class T, class SFINAE = decltype(*std::declval<T>())>
    constexpr bool is_indirection(bool)
    {
        return true;
    }
    template <class T>
    constexpr bool is_indirection(...)
    {
        return false;
    }
}
template <class T>
constexpr bool is_indirection()
{
    return detail::is_indirection<T>(true);
}

template <class T, bool ind = is_indirection<T>()>
struct underlying_type
{
    using type = T;
};

template <class T>
struct underlying_type<T, true>
{
    using type = typename std::remove_reference<decltype(*(std::declval<T>()))>::type;
};

template <class T>
typename std::enable_if<is_indirection<T>(), typename std::add_lvalue_reference<typename underlying_type<T>::type>::type>::type underlying_value(T&& val)
{
    return *std::forward<T>(val);
}

template <class T>
typename std::enable_if<!is_indirection<T>(), T&>::type underlying_value(T& val)
{
    return val;
}
template <class T>
typename std::enable_if<!is_indirection<T>(), const T&>::type underlying_value(const T& val)
{
    return val;
}


template <class T>
class Storage
{
public:
    T val;
    void print()
    {
        std::cout << underlying_value(val) << '\n';
    }
};

template <class T>
class StringStorage
{
public:
    T str;
    void printSize()
    {
        std::cout << underlying_value(str).size() << '\n';
    }
};

int main()
{
    int* a = new int(213);
    std::string str = "some string";
    std::shared_ptr<std::string> strPtr = std::make_shared<std::string>(str);
    Storage<int> sVal{ 1 };
    Storage<int*> sPtr{ a };
    Storage<std::string> sStrVal{ str };
    Storage<std::shared_ptr<std::string>> sStrPtr{ strPtr };
    StringStorage<std::string> ssStrVal{ str };
    StringStorage<const std::shared_ptr<std::string>> ssStrPtr{ strPtr };

    sVal.print();
    sPtr.print();
    sStrVal.print();
    sStrPtr.print();
    ssStrVal.printSize();
    ssStrPtr.printSize();

    std::cout << is_indirection<int*>() << '\n';
    std::cout << is_indirection<int>() << '\n';
    std::cout << is_indirection<std::shared_ptr<int>>() << '\n';
    std::cout << is_indirection<std::string>() << '\n';
    std::cout << is_indirection<std::unique_ptr<std::string>>() << '\n';
}