enable_if with copy constructors

问题

I am trying std::enable_if for the first time and struggling. Any guidance would be appreciated.

As a toy example, here is a simple static vector class, for which I want to define a copy constructor, but the behaviour depends on the relative sizes of the vectors:

just copy data into a smaller or same-sized vector
copy data into a larger vector and then pad the rest with zeroes

So the vector class is:

template <size_t _Size>
class Vector
{
    double _data[_Size];

public:
    Vector()
    {
        std::fill(_data, _data + _Size, 0.0);
    }

    const double* data() const
    {
        return _data;
    }
...
};

The copy constructor should support something like this, copying the first 2 elements of v3 into v2:

Vector<3> v3;
Vector<2> v2(v3);

I tried a copy constructor for behaviour 1. like this, which compiles:

template <size_t _OtherSize,
    typename = typename std::enable_if_t<_Size <= _OtherSize>>
    Vector(const Vector<_OtherSize>& v) : Vector()
    {
        std::copy(v.data(), v.data() + _Size, _data);
    }

but the compiler cannot distinguish this from behaviour 2. even though the enable_if conditions are mutually exclusive.

template <size_t _OtherSize,
    typename = typename std::enable_if_t<_OtherSize < _Size>>
    Vector(const Vector<_OtherSize>& v) : Vector()
    {
        std::copy(v.data(), v.data() + _OtherSize, _data);
        std::fill(_data + _OtherSize, _data + _Size, 0.0);
    }

I also tried putting enable_if in the argument instead, but it couldn't deduce the value of _OtherSize:

template <size_t _OtherSize>
    Vector(const typename std::enable_if_t<_Size <= _OtherSize, 
    Vector<_OtherSize>> & v)
    : Vector()
    {
        std::copy(v.data(), v.data() + _Size, _data);
    }

What is the best way to do this (using enable_if, not a simple if statement)?

Thanks

回答1:

Ignoring defaults, the signature of both of those constructors is

template <size_t N, typename>
Vector(const Vector<N>&)

I.e., they're ultimately the same.

One way to differentiate them is to make the template parameter type directly dependent on enable_if's condition:

template <size_t _OtherSize,
    std::enable_if_t<(_Size <= _OtherSize), int> = 0>
    Vector(const Vector<_OtherSize>& v) : Vector()
    {
        std::copy(v.data(), v.data() + _Size, _data);
    }

template <size_t _OtherSize,
    std::enable_if_t<(_OtherSize < _Size), int> = 0>
    Vector(const Vector<_OtherSize>& v) : Vector()
    {
        std::copy(v.data(), v.data() + _OtherSize, _data);
        std::fill(_data + _OtherSize, _data + _Size, 0.0);
    }

As an aside, names like _Size and _OtherSize are reserved for the implementation and thus illegal for user code – lose the underscore and/or the capital letter.

Also, as @StoryTeller hinted at, you don't want the first constructor to apply when _OtherSize == _Size, as the compiler-generated copy constructor has ideal behavior. Said constructor is already less specialized than the copy constructor for same-sized Vectors, so it won't be selected during overload resolution anyway, but it would be best to make the intent clear by switching <= to <.

回答2:

Don't use _Cap; they are reserved for the implementation. In fact, std source uses tgese names because they are reserved. Do not mimic std/system header internal naming conventions.

template <size_t O>
Vector(const Vector<O>& v) : Vector()
{
  constexpr auto to_copy = (std::min)( O, Size );
  constexpr auto to_fill = Size-to_copy;
  auto const* src=v.data();
  std::copy(src, src + to_copy, _data);
  std::fill(_data + to_copy, _data + to_copy+to_fill, 0.0);
}
Vector(const Vector& v) = default;

you'll find this optimizes down to the code you want; in the no-fill case, std::fill is called with (foo, foo, 0.0), and thr body is a loop that is easy to prove is a noop by a compiler when passed the same pointer twice.

来源：https://stackoverflow.com/questions/46403866/enable-if-with-copy-constructors

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c++

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sfinae

enable-if