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c++ - constexpr 查找实现

转载 作者:塔克拉玛干 更新时间:2023-11-03 00:47:31 24 4
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回答后this question和阅读 this talk看着 this code , 我想实现 constexpr find只有简单的数组类。

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, "");
}

compiles as expected

并使用自定义 constexpr 迭代器:

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;

这是唯一的区别,编译器给我错误:

in constexpr expansion of constexpr_find<array_iterator<const int>, int>(array.array<T,
N>::begin<int, 3u>(), array.array<T, N>::end<int, 3u>(), 0)

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

Live example

这是 gcc 错误吗,因为 clang 编译得很好,还是两个片段有区别?

最佳答案

我不能肯定地说,但是您将数组成员的指针存储到外部迭代器类中,这可能是该错误的原因。

--------更新开始--------

这是演示问题的最小片段:

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
}

似乎禁止在常量表达式中使用指向数组元素(具有非零偏移量)的指针。

--------更新结束--------

解决方法很简单 - 将指针存储到数组对象和偏移量。

Live

#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,可以实现启用 constexpr 的 C++11 版本的查找:

Live

#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");
}

您可能也有兴趣查看 Sprout library , 它包含大量的 constexpr 数据结构和算法。

关于c++ - constexpr 查找实现,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32397794/

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