实现指针元组的深度复制答案

【问题标题】：Implementing deep copying of tuple of pointers实现指针元组的深度复制
【发布时间】：2016-05-03 23:35:39
【问题描述】：

如果我错了，请纠正我，但是在运行时

std::tuple<double*, bool*> t(new double(3.5), new bool(true));
print_tuple(t);
std::tuple<double*, bool*> n = t;
print_tuple(n);

我明白了

std::get<0>(t) = 0x1f13d0
std::get<1>(t) = 0x1f13b0
std::get<0>(n) = 0x1f13d0
std::get<1>(n) = 0x1f13b0

这意味着元组中的指针只是浅拷贝的，对吗？因此，我编写了一个简单的实用程序，旨在深度复制元组中作为指针的所有元素：

template <std::size_t N, typename Tuple>
std::enable_if_t<std::is_pointer<std::tuple_element_t<N, Tuple>>::value> assign (Tuple& tuple, const Tuple& other) {
    std::get<N>(tuple) = new std::remove_pointer_t<std::tuple_element_t<N, Tuple>>(*std::get<N>(other));
}

template <std::size_t N, typename Tuple>
std::enable_if_t<!std::is_pointer<std::tuple_element_t<N, Tuple>>::value> assign (Tuple& tuple, const Tuple& other) {
    std::get<N>(tuple) = std::get<N>(other);
}

template <typename Tuple, std::size_t... Is>
Tuple deep_copy_impl (const Tuple& other, std::index_sequence<Is...>) {
    Tuple tuple = {};
    const int a[] = {(assign<Is>(tuple, other), 0)...};
    static_cast<void>(a);
    return tuple;
}

template <typename Tuple>
Tuple deep_copy (const Tuple& other) {
    return deep_copy_impl(other, std::make_index_sequence<std::tuple_size<Tuple>::value>{});
}

对于上面的示例，这似乎可以正常工作，但是当我使用由定义的元组 tup 尝试它时

std::tuple<double*, bool*> t(new double(3.5), new bool(true));
std::tuple<int*, std::tuple<double*, bool*>*, char> tup(new int(5), &t, 'a');

我得到了元组指针的深拷贝，但是元组指针中的指针又被浅拷贝了。我希望这些指针也能被深度复制。如何为任意数量的嵌套指针元组解决这个问题？这是我的测试结果：

#include <iostream>
#include <type_traits>
#include <utility>
#include <tuple>

template <std::size_t N, typename Tuple>
std::enable_if_t<std::is_pointer<std::tuple_element_t<N, Tuple>>::value> assign (Tuple& tuple, const Tuple& other) {
    std::get<N>(tuple) = new std::remove_pointer_t<std::tuple_element_t<N, Tuple>>(*std::get<N>(other));
}

template <std::size_t N, typename Tuple>
std::enable_if_t<!std::is_pointer<std::tuple_element_t<N, Tuple>>::value> assign (Tuple& tuple, const Tuple& other) {
    std::get<N>(tuple) = std::get<N>(other);
}

template <typename Tuple, std::size_t... Is>
Tuple deep_copy_impl (const Tuple& other, std::index_sequence<Is...>) {
    Tuple tuple = {};
    const int a[] = {(assign<Is>(tuple, other), 0)...};
    static_cast<void>(a);
    return tuple;
}

template <typename Tuple>
Tuple deep_copy (const Tuple& other) {
    return deep_copy_impl(other, std::make_index_sequence<std::tuple_size<Tuple>::value>{});
}

// Testing
template <typename Tuple, std::size_t... Is>
std::ostream& print_tuple_impl (const Tuple& tuple, std::ostream& os, std::index_sequence<Is...>) {
    const int a[] = {(os << "std::get<" << Is << ">(tuple) = " << std::get<Is>(tuple) << '\n', 0)...};
    static_cast<void>(a);
    return os;
}

template <typename Tuple>
std::ostream& print_tuple (const Tuple& tuple, std::ostream& os = std::cout) {
    return print_tuple_impl (tuple, os, std::make_index_sequence<std::tuple_size<Tuple>::value>{});
}

int main() {
    std::tuple<double*, bool*> t(new double(3.5), new bool(true));
    print_tuple(t);
    std::tuple<double*, bool*> n = t;
    print_tuple(n);
    std::cout << "Above is shallow copying only.\n\n";

    std::tuple<int*, std::tuple<double*, bool*>*, char> tup(new int(5), &t, 'a');
    print_tuple(tup);
    std::tuple<int*, std::tuple<double*, bool*>*, char> q = deep_copy(tup); 
    print_tuple(q);
    std::cout << "\nAbove seems like a deep copy, but look at this:\n";

    print_tuple(*std::get<1>(tup));
    print_tuple(*std::get<1>(q));
}

输出：

std::get<0>(tuple) = 0x1f13d0
std::get<1>(tuple) = 0x1f13b0
std::get<0>(tuple) = 0x1f13d0
std::get<1>(tuple) = 0x1f13b0
Above is shallow copying only.

std::get<0>(tuple) = 0x1f13f0
std::get<1>(tuple) = 0x72fe10
std::get<2>(tuple) = a
std::get<0>(tuple) = 0x1f1410
std::get<1>(tuple) = 0x1f1430
std::get<2>(tuple) = a

Above seems like a deep copy, but look at this:
std::get<0>(tuple) = 0x1f13d0
std::get<1>(tuple) = 0x1f13b0
std::get<0>(tuple) = 0x1f13d0
std::get<1>(tuple) = 0x1f13b0

【问题讨论】：

我对这个不太熟悉，但我怀疑如果当前的元组元素本身就是一个元组，你需要在deep_copy_impl中递归调用depp_copy。
啊！我会试试的。这可能需要结构的部分特化。
如果你能弄明白，请发布答案！我很想知道结果如何；）实际上 T.C.解决方案看起来很优雅。
@dau_sama 是的，T.C. 的解决方案非常优雅。您可能对我如何将 T.C. 的解决方案扩展到 STL 容器感兴趣。

标签： c++ tuples c++14 variadic-templates deep-copy

【解决方案1】：

template<class T> T deep_copy(const T& t);
template<class T> T* deep_copy(T* tp);
template<class... Ts> std::tuple<Ts...> deep_copy(const std::tuple<Ts...>&);

template<class T>
T deep_copy(const T& t) { return t; }

template<class T>
T* deep_copy(T* tp) { return new T(deep_copy(*tp)); }

template<class... Ts, size_t... Is>
std::tuple<Ts...> deep_copy_impl(const std::tuple<Ts...>& t, std::index_sequence<Is...>) {
    return std::tuple<Ts...>{deep_copy(std::get<Is>(t))... };
}

template<class... Ts>
std::tuple<Ts...> deep_copy(const std::tuple<Ts...>& t) {
    return deep_copy_impl(tuple, std::index_sequence_for<Ts...>());
}

【讨论】：

谢谢，一如既往。是否需要前 2 个前向声明？我测试了在我的具体示例中不需要它们，但在其他情况下是否需要它们？
@prestokeys 不是真的，但是很好地布置了整个重载集并且不必担心顺序。

【解决方案2】：

将 T.C. 的想法扩展到 STL 容器：

#include <iostream>
#include <type_traits>
#include <utility>
#include <tuple>
#include <vector>
#include <set>

template <typename T>
using void_t = void;

template <typename T, typename = void>
struct has_emplace_back : std::false_type {};

template <typename T>
struct has_emplace_back<T, void_t<decltype(std::declval<T>().emplace_back(std::declval<typename T::value_type>()))>> : std::true_type {};

template <typename T, typename = void>
struct has_emplace : std::false_type {};

template <typename T>
struct has_emplace<T, void_t<decltype(std::declval<T>().emplace(std::declval<typename T::value_type>()))>> : std::true_type {};
// etc... for other container types.

template <typename T>
struct is_stl_container : std::integral_constant<bool, has_emplace_back<T>::value || has_emplace<T>::value> {};

template <typename Container> std::enable_if_t<has_emplace_back<Container>::value, Container> deep_copy (const Container&);
template <typename Container> std::enable_if_t<has_emplace<Container>::value, Container> deep_copy (const Container&);
template <typename... Ts> std::tuple<Ts...> deep_copy (const std::tuple<Ts...>&);  // This forward declarations, though not needed to compile, is needed for the nested deep copying to work correctly.

template <typename T>
std::enable_if_t<!is_stl_container<T>::value, T> deep_copy (const T& t) {
    return t;
}

template <typename T>
T* deep_copy (T* t) {
    return new T(deep_copy(*t));  // Note that since T's copy constructor is called here, then if T is a custom class that has its own custom copy constructor that carries out deep copying, then t's pointer data members are also deep copied.
}

template <typename... Ts, std::size_t... Is>
std::tuple<Ts...> deep_copy_impl (const std::tuple<Ts...>& tuple, std::index_sequence<Is...>) {
    return std::tuple<Ts...>{ deep_copy(std::get<Is>(tuple))... };
}

template <typename... Ts>
std::tuple<Ts...> deep_copy (const std::tuple<Ts...>& tuple) {
    return deep_copy_impl(tuple, std::index_sequence_for<Ts...>{});
}

template <typename Container>
std::enable_if_t<has_emplace_back<Container>::value, Container> deep_copy (const Container& c) {
    Container container;
    for (const typename Container::value_type& t : c)
        container.emplace_back(deep_copy(t));
    return container;
}

template <typename Container>
std::enable_if_t<has_emplace<Container>::value, Container> deep_copy (const Container& c) {
    Container container;
    for (const typename Container::value_type& t : c)
        container.emplace(deep_copy(t));
    return container;
}

// Testing
template <typename Tuple, std::size_t... Is>
std::ostream& print_tuple_impl (const Tuple& tuple, std::ostream& os, std::index_sequence<Is...>) {
    const int a[] = {(os << "std::get<" << Is << ">(tuple) = " << std::get<Is>(tuple) << '\n', 0)...};
    static_cast<void>(a);
    return os;
}

template <typename Tuple>
std::ostream& print_tuple (const Tuple& tuple, std::ostream& os = std::cout) {
    return print_tuple_impl (tuple, os, std::make_index_sequence<std::tuple_size<Tuple>::value>{});
}

int main() {
    std::tuple<double*, bool*> t(new double(3.5), new bool(true));
    print_tuple(t);
    std::tuple<double*, bool*> n = t;
    print_tuple(n);
    std::cout << "Above is shallow copying only.\n\n";

    std::tuple<int*, std::tuple<double*, bool*>*, char> tup(new int(5), &t, 'a');
    print_tuple(tup);
    std::tuple<int*, std::tuple<double*, bool*>*, char> q = deep_copy(tup); 
    print_tuple(q);
    std::cout << "\nAbove and below show that we have full deep copying:\n";

    print_tuple(*std::get<1>(tup));
    print_tuple(*std::get<1>(q));

    std::cout << "\nDeep copy of a vector of pointers:\n";
    std::vector<int*> v = {new int(5), new int(2), new int(8)};
    for (int* x : v) std::cout << x << ' ';  std::cout << '\n';
    std::vector<int*> u = deep_copy(v);
    for (int* x : u) std::cout << x << ' ';  std::cout << '\n';

    std::cout << "\nDeep copy of a set of pointers:\n";
    std::set<int*> s = {new int(5), new int(2), new int(8)};
    for (int* x : s) std::cout << x << ' ';  std::cout << '\n';
    std::set<int*> ss = deep_copy(s);
    for (int* x : ss) std::cout << x << ' ';  std::cout << '\n';
}

输出：

std::get<0>(tuple) = 0x8513d0
std::get<1>(tuple) = 0x8513b0
std::get<0>(tuple) = 0x8513d0
std::get<1>(tuple) = 0x8513b0
Above is shallow copying only.

std::get<0>(tuple) = 0x8513f0
std::get<1>(tuple) = 0x72fd70
std::get<2>(tuple) = a
std::get<0>(tuple) = 0x851410
std::get<1>(tuple) = 0x851470
std::get<2>(tuple) = a

Above and below show that we have full deep copying:
std::get<0>(tuple) = 0x8513d0
std::get<1>(tuple) = 0x8513b0
std::get<0>(tuple) = 0x851430
std::get<1>(tuple) = 0x851450

Deep copy of a vector of pointers:
0x851490 0x8514b0 0x8514d0
0x851510 0x851550 0x851530

Deep copy of a set of pointers:
0x851570 0x8515c0 0x8515e0
0x851690 0x851800 0x851870

【讨论】：