计算嵌套包中的类型总数

Question

NumTypes<Args...>::value是给出Args...中类型的总数，包括嵌套包中的所有类型（如果有的话），例如如果

using T = Group<int, bool, Wrap<char, Pack<char, long, Group<char, Object, short>, short>, double>, long>;

然后 NumTypes<T, int, char>::value 将是 13（我们不计算包装器类本身）。下面的代码工作正常，但是当我用 std::string 替换任何类型时，我得到一连串的 std::allocator 错误，这些错误永远不会终止（使用 GCC 4.8.1）。我怀疑其他类型会产生相同的错误。为什么？以及如何修复代码以避免出现这种奇怪的错误？

#include <iostream>
#include <string>

#define show(variable) std::cout << #variable << " = " << variable << std::endl;

template <typename T>
struct IsPack {
    static const bool value = false;
};

template <template <typename...> class P, typename... Args>
struct IsPack<P<Args...>> {
    static const bool value = true;
};

template <typename...> struct NumTypes;

template <typename T>
struct NumTypes<T> {
    static const int value = 1;
};

template <template <typename...> class P>
struct NumTypes<P<>> { static const int value = 0; };

template <template <typename...> class P, typename First, typename... Rest>
struct NumTypes<P<First, Rest...>> {
    static const int value = IsPack<First>::value ?
        NumTypes<First>::value + NumTypes<P<Rest...>>::value :
        1 + NumTypes<P<Rest...>>::value;
};

template <typename First, typename... Rest>
struct NumTypes<First, Rest...> {
    static const int value = NumTypes<First>::value + NumTypes<Rest...>::value;
};

template <typename...> struct Pack;
template <typename...> struct Group;
template <typename...> struct Wrap;
struct Object {};

int main() {
    using A = Pack<int, Object, long>;
    show (NumTypes<A>::value)  // 3

    using B = Pack<int, bool, Pack<char, Object>, long>;
    show (NumTypes<B>::value)  // 5

    using C = Group<int, bool, Wrap<char, Pack<char, long, Group<char, Object, short>, short>, double>, long>;
    show (NumTypes<C>::value)  // 11

    using D = Group<Pack<int, Object, double>, bool, Wrap<char, Pack<char, double, Group<char, Pack<char, long, short>, int, Object>, short>, double>, long>;
    show (NumTypes<D>::value)  // 16

    std::cout << NumTypes<A, B, int, char, C, Object, D>::value << std::endl;  // 38
}

Answer 1

std::string 是 std::basic_string<char> 的 typedef，因此匹配 IsPack 和 NumTypes 的特化，用于分组类型。

您必须专注于分组或模板叶类型，在您的情况下，以更省力的方式为准：

template <typename T>
struct IsPack
    : std::false_type
{};


template <typename...> 
struct NumTypes;

template <typename T, bool is_pack> 
struct NumTypesHelper;

template <typename T>
struct NumTypesHelper<T, false>
    : std::integral_constant<int, 1>
{};

template <template <typename...> class P, typename... Args>
struct NumTypesHelper<P<Args...>, true>
    : NumTypes<Args...>
{};

template <>
struct NumTypes<>
    : std::integral_constant<int, 0>
{};

template <typename First, typename... Rest>
struct NumTypes<First, Rest...>
    : std::integral_constant<int, NumTypesHelper<First, IsPack<First>::value>::value + NumTypes<Rest...>::value>
{};

template <typename...> struct Pack;
template <typename...> struct Group;
template <typename...> struct Wrap;
struct Object {};

template <typename... Args>
struct IsPack<Pack<Args...>>
    : std::true_type
{};

template <typename... Args>
struct IsPack<Group<Args...>>
    : std::true_type
{};

template <typename... Args>
struct IsPack<Wrap<Args...>>
    : std::true_type
{};

#define show(variable) std::cout << #variable << " = " << variable << std::endl;

int main() {
    using A = Pack<int, Object, long>;
    show (NumTypes<A>::value)  // 3

    using B = Pack<int, bool, Pack<char, double>, long>;
    show (NumTypes<B>::value)  // 5

    using C = Group<int, bool, Wrap<char, Pack<char, long, Group<char, Object, short>, short>, double>, long>;
    show (NumTypes<C>::value)  // 11

    using D = Group<Pack<int, Object, double>, bool, Wrap<char, Pack<char, double, Group<char, Pack<char, long, short>, int, Object>, short>, double>, long>;
    show (NumTypes<D>::value)  // 16
}

Answer 2

std::string 实际上是 std::basic_string<char, char_traits<char>, allocator<charT>>.
它的声明有默认的模板参数：

template <class charT,
          class traits = char_traits<charT>,
          class Alloc = allocator<charT>>
class basic_string;

你的递归是如此错误，而且你有无限循环（当你从 basic_string<charT, traits, Alloc> 中删除第一个参数时，你得到 basic_string<traits, Alloc, allocator<traits>>）。

您可以通过从等式中删除 P 来解决这个问题：

template <template <typename...> class P, typename First, typename... Rest>
struct NumTypes<P<First, Rest...>> {
    static const int numInFirst = NumTypes<First>::value;
    static const int value = numInFirst + NumTypes<std::tuple<Rest...>>::value;
};

Live example

计算嵌套包中的类型总数

Counting the total number of types in a nested pack

c++

templates

variadic

c++11