Constexpr查找实现 [英] Constexpr find implementation

问题描述

回答此问题后，阅读本演讲并查看此代码，我想实现 constexpr使用简单的数组类查找。

After answering this question and reading this talk and looking at this code, I want to implement constexpr find with just simple array class.

请考虑以下示例：

#include <cstddef>

template <class It, class T>
constexpr auto constexpr_find(const It& b, const It& e, T value) {
    auto begin = b;
    while (begin != e) {
        if (*begin == value) break;

        ++begin;
    }
    return *begin;
}

template<typename T, size_t N>
class array
{
public:
   typedef T* iterator;
   typedef const T* const_iterator;
   constexpr auto begin() const { return const_iterator(array_); }
   constexpr auto end() const { return const_iterator(array_ + N); }

   T array_[N];
   static constexpr size_t size = N;
};

int main()
{
   constexpr array<int, 3> array{{0,2,3}};
   static_assert(constexpr_find(array.begin(), array.end(), 0) == 0, "");
}

按预期编译

并使用自定义constexpr迭代器：

And with custom constexpr iterator:

template<class T>
class array_iterator
{
public:
   constexpr array_iterator(const T* v) : iterator(v)
   {
   }
   constexpr const T& operator * () const { return *iterator; }
   constexpr array_iterator& operator ++()
   {
      ++iterator;
      return *this;
   }
   constexpr bool operator != (const array_iterator& other) const { return iterator != other.iterator; }
private:
   const T* iterator;
};

在数组类中：

typedef const array_iterator<const T> const_iterator;

唯一的区别是，编译器给我错误：

that's the only difference, compiler give me error:

在
constexpr_find< array_iterator< const int> ;, int>（array.array< T，
N> ;: begin< int，3u>（），array.array< T，N> :: end< int，3u>（），0）

错误：（（（（const int *）（& array.array< int，3u> :: array_）））！=（（（（const
int *）（& array .array< int，3u> :: array_））+ 12u））不是常数
表达式

error: (((const int*)(& array.array<int, 3u>::array_)) != (((const int*)(& array.array<int, 3u>::array_)) + 12u)) is not a constant expression

在线示例

这是gcc错误，因为用clang编译得很好，还是两个代码段之间存在差异？

Is this gcc bug, since clang compile this fine, or there is difference in two snippets?

推荐答案

我不能说当然，但是您将数组成员的指针存储到外部迭代器类中，这可能是导致该错误的原因。

I cannot say for sure, but you store pointers for array's member into external iterator class, it may be the reason for that error.

---------更新开始---------

--------- update start ---------

这是演示该问题的最小代码段：

Here is the minimal snippet that demonstrates the problem:

constexpr const struct A { int i[2]; } a {{0,0}};

int main ()
{
  static_assert (nullptr != a.i  , ""); // ok
  static_assert (nullptr != a.i+0, ""); // ok
  static_assert (nullptr != a.i+1, ""); // error
}

似乎禁止使用指向数组元素的指针（带有

It seems to be forbidden to have pointers to array elements (with non-zero offset) in constant expressions.

---------更新结束---------

--------- update end ---------

解决方法很简单-存储指向数组对象和偏移量的指针。

The workaround is trivial - store the pointer to array object and offset.

实时

#include <cstddef>

template <class It, class T>
constexpr auto constexpr_find(const It& b, const It& e, T value) {
    auto begin = b, end = e;
    while (begin != end) {
        if (*begin == value) break;

        ++begin;
    }
    return *begin;
}

template<class Array>
class array_iterator
{
public:
   constexpr array_iterator(const Array& a, size_t pos=0u) : array_(&a), pos_ (pos)
   {
   }
   constexpr const typename Array::value_type& 
   operator * () const { return (*array_)[pos_]; }

    constexpr array_iterator& operator ++()
   {
      ++pos_;
      return *this;
   }
   constexpr bool operator != (const array_iterator& other) const 
   { return array_ != other.array_ || pos_ != other.pos_; }

private:
   const Array* array_;
   size_t pos_;
};

template<typename T, size_t N>
class array
{
public:
   typedef T value_type;
   typedef const array_iterator<array> const_iterator;
   constexpr T const& operator[] (size_t idx) const { return array_[idx]; }
   constexpr auto begin() const { return const_iterator(*this); }
   constexpr auto end() const { return const_iterator(*this, N); }

   T array_[N];
   static constexpr size_t size = N;
};

int main()
{
   constexpr array<int, 3> array{{0,2,3}};
   static_assert(constexpr_find(array.begin(), array.end(), 0) == 0, "");
}

顺便说一句，可以实现constexpr的C ++ 11版本启用的查找：

By the way, it is possible to implement C++11 version of constexpr enabled find:

在线

#include <cstddef>
#include <cassert>

#if !defined(__clang__) && __GNUC__ < 5
// TODO: constexpr asserts does not work in gcc4, but we may use 
// "thow" workaround from 
// http://ericniebler.com/2014/09/27/assert-and-constexpr-in-cxx11/
# define ce_assert(x) ((void)0)
#else
# define ce_assert(x) assert(x)
#endif
namespace my {

template <class It, class T>
inline constexpr It
find (It begin, It end, T const& value) noexcept
{
    return ! (begin != end && *begin != value)
         ? begin
         : find (begin+1, end, value);
}

template<class Array>
class array_iterator
{
public:
  using value_type = typename Array::value_type;

  constexpr array_iterator(const Array& array, size_t size = 0u) noexcept
    : array_ (&array)
    , pos_ (size)
  {}

  constexpr const value_type operator* () const noexcept
  {
    return ce_assert (pos_ < Array::size), (*array_) [pos_];
  }

#if __cplusplus >= 201402L // C++14
  constexpr
#endif
  array_iterator& operator ++() noexcept
  {
    return ce_assert (pos_ < Array::size), ++pos_, *this;
  }

  constexpr array_iterator operator+ (size_t n) const noexcept
  {
    return ce_assert (pos_+n <= Array::size), array_iterator (*array_, pos_+n);
  }

  friend constexpr bool
  operator != (const array_iterator& i1, const array_iterator& i2) noexcept
  {
    return i1.array_ != i2.array_ || i1.pos_ != i2.pos_;
  }

  friend constexpr size_t
  operator- (array_iterator const& i1, array_iterator const& i2) noexcept
  {
    return ce_assert (i1.array_ == i2.array_), i1.pos_ - i2.pos_;
  }

private:
  const Array* array_;
  size_t pos_;
};

template<typename T, size_t N>
class array
{
public:
  using value_type = T;
  using const_iterator = const array_iterator<array>;

  constexpr value_type const&
  operator[] (size_t idx) const noexcept
  { return array_[idx];  }

  constexpr const_iterator begin() const noexcept
  { return const_iterator(*this); }

  constexpr const_iterator end()   const noexcept
  { return const_iterator(*this, N); }

  T array_[N];
  static constexpr size_t size = N;
};

}

int main()
{
  static constexpr my::array<int, 3> array{{0,2,3}};

  static_assert (
    find (array.begin(), array.end(), 2) - array.begin () == 1,
    "error");
}

您可能还想检查 Sprout库，它包含许多constexpr数据结构和算法。

You may also be interested to check Sprout library, it contains a lot of constexpr data structures and algorithms.

这篇关于Constexpr查找实现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！