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[英]How to initialize an std::array<T, 2> where T is non-copyable and non-default-constructible?
[英]How to initialize std::array<T, n> elegantly if T is not default constructible?
如果 T 不是默认可构造的,如何初始化std::array<T, n>
?
我知道可以像这样初始化它:
T t{args};
std::array<T, 5> a{t, t, t, t, t};
但对我来说n
是模板参数:
template<typename T, int N>
void f(T value)
{
std::array<T, N> items = ???
}
即使它不是模板,如果n
太大,手动重复值也是相当难看的。
给定 N,您可以使用名为genseq_t<>
的生成器生成名为seq<0,1,2,3,...N-1>
的序列类型,然后执行以下操作:
template<typename T, int N>
void f(T value)
{
//genseq_t<N> is seq<0,1,...N-1>
std::array<T, N> items = repeat(value, genseq_t<N>{});
}
其中repeat
定义为:
template<typename T, int...N>
auto repeat(T value, seq<N...>) -> std::array<T, sizeof...(N)>
{
//unpack N, repeating `value` sizeof...(N) times
//note that (X, value) evaluates to value
return {(N, value)...};
}
其余的定义为:
template<int ... N>
struct seq
{
using type = seq<N...>;
static const std::size_t size = sizeof ... (N);
template<int I>
struct push_back : seq<N..., I> {};
};
template<int N>
struct genseq : genseq<N-1>::type::template push_back<N-1> {};
template<>
struct genseq<0> : seq<> {};
template<int N>
using genseq_t = typename genseq<N>::type;
希望有帮助。
遗憾的是,这里的现有答案不适用于不可复制的类型。 所以我接受了@Nawaz 的回答并对其进行了修改:
#include <utility>
#include <array>
template<typename T, size_t...Ix, typename... Args>
std::array<T, sizeof...(Ix)> repeat(std::index_sequence<Ix...>, Args &&... args) {
return {{((void)Ix, T(args...))...}};
}
template<typename T, size_t N>
class initialized_array: public std::array<T, N> {
public:
template<typename... Args>
initialized_array(Args &&... args)
: std::array<T, N>(repeat<T>(std::make_index_sequence<N>(), std::forward<Args>(args)...)) {}
};
请注意,这是一个std::array
子类,因此可以轻松编写
class A {
A(int, char) {}
}
...
class C {
initialized_array<A, 5> data;
...
C(): data(1, 'a') {}
}
无需重复类型和大小。 当然,这种方式也可以实现为函数initialize_array
。
以下将解决您的问题:
#if 1 // Not in C++11, but in C++1y (with a non linear better version)
template <std::size_t ...> struct index_sequence {};
template <std::size_t I, std::size_t ...Is>
struct make_index_sequence : make_index_sequence<I - 1, I - 1, Is...> {};
template <std::size_t ... Is>
struct make_index_sequence<0, Is...> : index_sequence<Is...> {};
#endif
namespace detail
{
template <typename T, std::size_t ... Is>
constexpr std::array<T, sizeof...(Is)>
create_array(T value, index_sequence<Is...>)
{
// cast Is to void to remove the warning: unused value
return {{(static_cast<void>(Is), value)...}};
}
}
template <std::size_t N, typename T>
constexpr std::array<T, N> create_array(const T& value)
{
return detail::create_array(value, make_index_sequence<N>());
}
所以测试一下:
struct NoDefaultConstructible {
constexpr NoDefaultConstructible(int i) : m_i(i) { }
int m_i;
};
int main()
{
constexpr auto ar1 = create_array<10>(NoDefaultConstructible(42));
constexpr std::array<NoDefaultConstructible, 10> ar2 = create_array<10>(NoDefaultConstructible(42));
return 0;
}
受@Nawaz 启发,这更简洁,更好地利用了标准: https ://godbolt.org/z/d6a7eq8T5
#include <array>
#include <type_traits>
#include <fmt/format.h>
#include <fmt/ranges.h>
// Inspired by https://stackoverflow.com/a/18497366/874660
//! Return a std::array<T, Sz> filled with
//! calls to fn(i) for i in the integer sequence:
template <typename Int, Int...N, typename Fn>
[[nodiscard]] constexpr auto transform_to_array(std::integer_sequence<Int, N...>, Fn fn) {
return std::array{(static_cast<void>(N), fn(N))...};
}
//! Repeated application of nullary fn:
template <std::size_t N, typename Fn>
[[nodiscard]] constexpr auto generate_n_to_array(Fn fn) -> std::array<decltype(fn()), N> {
return transform_to_array(std::make_integer_sequence<std::size_t, N>(), [&](std::size_t) { return fn(); });
}
int main() {
static constexpr std::array<int, 3> a = generate_n_to_array<3>([i = 0]() mutable { return 2 * (i++); });
fmt::print("{}\n", a);
}
已经有很多很好的答案,所以一句话:他们中的大多数人使用N
两次:一次在 array<T,N> 中,一次在 repeat/genseq/generate/whatever you call it...
由于无论如何我们都使用单个值,看来我们可以避免重复大小两次,方法如下:
#include <iostream>
#include <array>
#include <type_traits>
// Fill with a value -- won't work if apart from non-default-constructible, the type is also non-copyable...
template <typename T>
struct fill_with {
T fill;
constexpr fill_with(T value) : fill{value} {}
template <typename... Args>
constexpr fill_with(std::in_place_type_t<T>, Args&&... args) : fill{ T{std::forward<Args>(args)...} } {}
template <typename U, size_t N>
constexpr operator std::array<U, N> () {
return [&]<size_t... Is>(std::index_sequence<Is...>) {
return std::array{ ((void)Is, fill)... };
}(std::make_index_sequence<N>{});
}
};
// A little more generic, but requires C++17 deduction guides, otherwise using it with lambdas is a bit tedious
template <typename Generator>
struct construct_with {
Generator gen;
template <typename F> constexpr construct_with(F && f) : gen{std::forward<F>(f)} {}
template <typename U, size_t N>
constexpr operator std::array<U, N> () {
return [&]<size_t... Is>(std::index_sequence<Is...>) {
return std::array{ ((void)Is, gen())... };
}(std::make_index_sequence<N>{});
}
};
template <typename F>
construct_with(F&&) -> construct_with<F>;
struct A {
A(int){}
A(A const&) = delete;
};
int main() {
std::array<int, 7> arr = fill_with<int>{ 7 };
std::array<int, 7> arr = fill_with{ 7 }; // Ok since C++17
// or you can create a maker function to wrap it in pre-c++17 code, as usual.
std::array<A, 7> as = construct_with{[]{ return A{7}; }};
for (auto & elem : arr) std::cout << elem << '\n';
}
注意:在这里,我使用了一些 C++20 结构,在您的情况下这可能是问题,也可能不是问题,但是从一个更改为另一个非常简单
如果可能的话,我还建议使用 <type_traits> 和 std::index_sequence 以及 std 中的所有其他sequence
,以避免重新发明轮子)
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