通用转换运算符模板和移动语义：任何通用解决方案？ [英] Generic conversion operator templates and move semantics: any universal solution?

问题描述

这是操作中的显式引用限定转化操作符模板的后续操作。我已经尝试了许多不同的选项，我在这里给出一些结果，试图看看是否有任何解决方案最终。

This is a follow-up of Explicit ref-qualified conversion operator templates in action. I have experimented with many different options and I am giving some results here in an attempt to see if there is any solution eventually.

说一个类（例如 any ）需要提供转换到任何可能的类型在一个方便，安全（没有惊喜）的方式，保留移动语义。我可以想出四种不同的方式。

Say a class (e.g. any) needs to provide conversion to any possible type in a convenient, safe (no surprises) way that preserves move semantics. I can think of four different ways.

struct A
{
    // explicit conversion operators (nice, safe?)
    template<typename T> explicit operator T&&       () &&;
    template<typename T> explicit operator T&        () &;
    template<typename T> explicit operator const T&  () const&;

    // explicit member function (ugly, safe)
    template<typename T> T&&       cast() &&;
    template<typename T> T&        cast() &;
    template<typename T> const T&  cast() const&;
};

// explicit non-member function (ugly, safe)
template<typename T> T&&       cast(A&&);
template<typename T> T&        cast(A&);
template<typename T> const T&  cast(const A&);

struct B
{
    // implicit conversion operators (nice, dangerous)
    template<typename T> operator T&&       () &&;
    template<typename T> operator T&        () &;
    template<typename T> operator const T&  () const&;
};

最有问题的情况是初始化对象或对对象的右值引用，右值引用。函数调用工作在所有情况下（我想），但我发现他们太详细：

The most problematic cases are to initialize an object or an rvalue reference to an object, given a temporary or an rvalue reference. Function calls work in all cases (I think) but I find them too verbose:

A a;
B b;

struct C {};

C member_move = std::move(a).cast<C>();  // U1. (ugly) OK
C member_temp = A{}.cast<C>();           // (same)

C non_member_move(cast<C>(std::move(a)));  // U2. (ugly) OK
C non_member_temp(cast<C>(A{}));           // (same)

因此，我接下来尝试使用转换运算符：

So, I next experiment with conversion operators:

C direct_move_expl(std::move(a));  // 1. call to constructor of C ambiguous
C direct_temp_expl(A{});           // (same)

C direct_move_impl(std::move(b));  // 2. call to constructor of C ambiguous
C direct_temp_impl(B{});           // (same)

C copy_move_expl = std::move(a);  // 3. no viable conversion from A to C
C copy_temp_expl = A{};           // (same)

C copy_move_impl = std::move(b);  // 4. OK
C copy_temp_impl = B{};           // (same)

看来， const& overload可以在一个右值上调用，这会产生歧义，使得隐式转换的副本初始化作为唯一选项。

It appears that the const& overload is callable on an rvalue, which gives ambiguities, leaving copy-initialization with an implicit conversion as the only option.

但是， ：

template<typename T>
struct flexi
{
    static constexpr bool all() { return true; }

    template<typename A, typename... B>
    static constexpr bool all(A a, B... b) { return a && all(b...); }

    template<typename... A>
    using convert_only = typename std::enable_if<
        all(std::is_convertible<A, T>{}...),
    int>::type;

    template<typename... A>
    using explicit_only = typename std::enable_if<
        !all(std::is_convertible<A, T>{}...) &&
        all(std::is_constructible<T, A>{}...),
    int>::type;

    template<typename... A, convert_only<A...> = 0>
    flexi(A&&...);

    template<typename... A, explicit_only<A...> = 0>
    explicit flexi(A&&...);
};

using D = flexi<int>;

它提供了一般的隐式或显式构造函数，这取决于输入参数是隐式还是显式转换为某种类型。这种逻辑不是那么奇怪，例如。一些实现 std :: tuple 可以这样。现在，初始化 D 会产生

which provides generic implicit or explicit constructors depending on whether the input arguments can be implicitly or explicitly converted to a certain type. Such logic is not that exotic, e.g. some implementation of std::tuple can be like that. Now, initializing a D gives

D direct_move_expl_flexi(std::move(a));  // F1. call to constructor of D ambiguous
D direct_temp_expl_flexi(A{});           // (same)

D direct_move_impl_flexi(std::move(b));  // F2. OK
D direct_temp_impl_flexi(B{});           // (same)

D copy_move_expl_flexi = std::move(a);  // F3. no viable conversion from A to D
D copy_temp_expl_flexi = A{};           // (same)

D copy_move_impl_flexi = std::move(b);  // F4. conversion from B to D ambiguous
D copy_temp_impl_flexi = B{};           // (same)

出于不同的原因，唯一可用的选项direct-initialization带有隐式转换。但是，这正是隐式转换危险的地方。 b 可能实际上包含 D ，这可能是一种容器，但工作组合调用 D 的构造函数作为完全匹配，其中 b 表现为容器的一个假的元素导致运行时错误或灾难。

For different reasons, the only available option direct-initialization with an implicit conversion. However, this is exactly where implicit conversion is dangerous. b might actually contain a D, which may be a kind of container, yet the working combination is invoking D's constructor as an exact match, where b behaves like a fake element of the container, causing a runtime error or disaster.

最后，让我们尝试初始化一个右值引用：

Finally, let's try to initialize an rvalue reference:

D&& ref_direct_move_expl_flexi(std::move(a));  // R1. OK
D&& ref_direct_temp_expl_flexi(A{});           // (same)

D&& ref_direct_move_impl_flexi(std::move(b));  // R2. initialization of D&& from B ambiguous
D&& ref_direct_temp_impl_flexi(B{});           // (same)

D&& ref_copy_move_expl_flexi(std::move(a));  // R3. OK
D&& ref_copy_temp_expl_flexi(A{});           // (same)

D&& ref_copy_move_impl_flexi = std::move(b);  // R4. initialization of D&& from B ambiguous
D&& ref_copy_temp_impl_flexi = B{};           // (same)

看起来每个用例都有自己的要求，可能会在所有情况下工作。

It appears that every use case has its own requirements and there is no combination that might work in all cases.

更糟糕的是，以上所有结果都是clang 3.3;其他编译器和版本给出略有不同的结果，再没有通用解决方案。例如：实例。

What's worse, all above results are with clang 3.3; other compilers and versions give slightly different results, again with no universal solution. For instance: live example.

因此： strong>有什么可能的东西可能按照需要工作，或者我应该放弃转换操作符和坚持显式函数调用？

So: is there any chance something might work as desired or should I give up conversion operators and stick with explicit function calls?

推荐答案

不幸的是，C ++标准没有任何特殊的规则来解决这个特定的歧义。问题来自于你试图重载两个不同的事情：编译器试图转换的类型;

The C++ standard unfortunately does not have any special rule to resolve this particular ambiguity. The problem come from the fact that you are trying to overload on 2 different things: the type that the compiler is trying to convert to; and the kind of reference from which you are trying to convert from.

通过引入代理类，您可以在两个步骤中拆分分辨率。步骤1：确定它是r值引用，l值引用还是const l值引用。步骤2：转换为任何类型，保持步骤1中关于参考种类的决定。这样，你可以使用带有cast（）函数的解决方案，但是不必指定类型：

By introducing proxy classes, you can split the resolution in 2 steps. Step 1: decide if it's an r-value reference, an l-value reference, or a const l-value reference. Step 2: convert to any type, keeping the decision made in step 1 about the kind of reference. That way, you can use your solution with a cast() function but save you from having to specify the type:

struct A
{
    class A_r_ref
    {
        A* a_;
    public:
        A_r_ref(A* a) : a_(a) {}
        template <typename T> operator T&&() const&&;
    };

    struct A_ref
    {
        A* a_;
    public:
        A_ref(A* a) : a_(a) {}
        template <typename T> operator T&() const&&;
    };

    struct A_const_ref
    {
        const A* a_;
    public:
        A_const_ref(const A* a) : a_(a) {}
        template <typename T> operator const T&() const&&;
    };

    A_r_ref cast() && { return A_r_ref(this); }
    A_ref cast() & { return A_ref(this); }
    A_const_ref cast() const& { return A_const_ref(this); }
};

这篇关于通用转换运算符模板和移动语义：任何通用解决方案？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！