c++ - constexpr 连接两个或多个 char 字符串

Question

我正在尝试创建一个 constexpr 函数，该函数将连接任意数量的 char 数组，方法是根据 Xeo 的以下答案进行操作，该答案连接两个 char 数组。

https://stackoverflow.com/a/13294458/1128289

#include <array>

template<unsigned... Is> struct seq{};
template<unsigned N, unsigned... Is>
struct gen_seq : gen_seq<N-1, N-1, Is...>{};
template<unsigned... Is>
struct gen_seq<0, Is...> : seq<Is...>{};

template<unsigned N1, unsigned... I1, unsigned N2, unsigned... I2>
constexpr std::array<char const, N1+N2-1> concat(char const (&a1)[N1], char const (&a2)[N2], seq<I1...>, seq<I2...>){
  return {{ a1[I1]..., a2[I2]... }};
}

template<unsigned N1, unsigned N2>
constexpr std::array<char const, N1+N2-1> concat(char const (&a1)[N1], char const (&a2)[N2]){
  return concat(a1, a2, gen_seq<N1-1>{}, gen_seq<N2>{});
}

到目前为止我的尝试:

#include <iostream>
#include <array>

template<unsigned... Is> struct seq{};
template<unsigned N, unsigned... Is>
struct gen_seq : gen_seq<N-1, N-1, Is...>{};
template<unsigned... Is>
struct gen_seq<0, Is...> : seq<Is...>{};

template<unsigned N1, unsigned... I1, unsigned N2, unsigned... I2>
constexpr const std::array<char, N1+N2-1>
concat_impl(
    const char (&a1)[N1], const char (&a2)[N2], seq<I1...>, seq<I2...>)
{
    return {{ a1[I1]..., a2[I2]... }};
}

template<unsigned N1, unsigned N2>
constexpr const std::array<char, N1+N2-1>
concat(const char (&a1)[N1], const char (&a2)[N2])
{
    return concat_impl(a1, a2, gen_seq<N1-1>{}, gen_seq<N2>{});
}

template<unsigned N1, unsigned N2, class... Us>
constexpr auto
concat(const char(&a1)[N1], const char(&a2)[N2], const Us&... xs)
-> std::array<char, N1 + decltype(concat(a2, xs...))::size() - 1>
{
    return concat(a1, concat(a2, xs...));
}

int main()
{
    auto const s = concat("hi ", "there!");
    std::cout << s.data() << std::endl;
    // compile error:
    auto const t = concat("hi ", "there ", "how ", "are ", "you?");
    std::cout << t.data() << std::endl;
}

gcc 4.9 和 clang 3.5 都给出错误，表明找不到与 decltype 表达式中的 concat 匹配的函数。

clang :

error: no matching function for call to 'concat'
    auto const t = concat("hi ", "there ", "how ", "are ", "you?");
                   ^~~~~~
ctconcat.cpp:105:16: note: candidate template ignored: substitution failure [with N1 = 4, N2 = 7, Us = <char [5], char [5], char [5]>]: no matching function for call to 'concat'
constexpr auto concat(const char(&a1)[N1], const char(&a2)[N2], const Us&... xs) -> std::array<char, N1 + decltype(concat(a2, xs...))::size() - 1>
               ^                                                                                                   ~~~~~~
ctconcat.cpp:62:43: note: candidate function template not viable: requires 2 arguments, but 5 were provided
constexpr const std::array<char, N1+N2-1> concat(const char (&a1)[N1], const char (&a2)[N2])
                                          ^
1 error generated.

gcc 和 clang 的错误都表明第二个 concat 函数模板不是 decltype 表达式中的 concat 的候选者。仅考虑第一个模板。为什么会这样？我该如何解决？

编辑:关于为什么decltype不能递归使用的相关问题

trailing return type using decltype with a variadic template function

Answer 1

template<size_t S>
using size=std::integral_constant<size_t, S>;

template<class T, size_t N>
constexpr size<N> length( T const(&)[N] ) { return {}; }
template<class T, size_t N>
constexpr size<N> length( std::array<T, N> const& ) { return {}; }

template<class T>
using length_t = decltype(length(std::declval<T>()));
constexpr size_t sum_string_sizes() { return 0; }
template<class...Ts>
constexpr size_t sum_string_sizes( size_t i, Ts... ts ) {
  return (i?i-1:0) + sum_sizes(ts...);
}

然后

template
template<unsigned N1, unsigned N2, class... Us>
constexpr auto
concat(const char(&a1)[N1], const char(&a2)[N2], const Us&... xs)
-> std::array<char, sum_string_sizes( N1, N2, length_t<Us>::value... )+1 >
{
  return concat(a1, concat(a2, xs...));
}

它摆脱了 decltype 中的递归。

---

这是使用上述方法的完整示例:

template<size_t S>
using size=std::integral_constant<size_t, S>;

template<class T, size_t N>
constexpr size<N> length( T const(&)[N] ) { return {}; }
template<class T, size_t N>
constexpr size<N> length( std::array<T, N> const& ) { return {}; }

template<class T>
using length_t = decltype(length(std::declval<T>()));

constexpr size_t string_size() { return 0; }
template<class...Ts>
constexpr size_t string_size( size_t i, Ts... ts ) {
  return (i?i-1:0) + string_size(ts...);
}
template<class...Ts>
using string_length=size< string_size( length_t<Ts>{}... )>;

template<class...Ts>
using combined_string = std::array<char, string_length<Ts...>{}+1>;

template<class Lhs, class Rhs, unsigned...I1, unsigned...I2>
constexpr const combined_string<Lhs,Rhs>
concat_impl( Lhs const& lhs, Rhs const& rhs, seq<I1...>, seq<I2...>)
{
  // the '\0' adds to symmetry:
  return {{ lhs[I1]..., rhs[I2]..., '\0' }};
}

template<class Lhs, class Rhs>
constexpr const combined_string<Lhs,Rhs>
concat(Lhs const& lhs, Rhs const& rhs)
{
  return concat_impl(
    lhs, rhs,
    gen_seq<string_length<Lhs>{}>{},
    gen_seq<string_length<Rhs>{}>{}
 );
}

template<class T0, class T1, class... Ts>
constexpr const combined_string<T0, T1, Ts...>
concat(T0 const&t0, T1 const&t1, Ts const&...ts)
{
  return concat(t0, concat(t1, ts...));
}

template<class T>
constexpr const combined_string<T>
concat(T const&t) {
  return concat(t, "");
}
constexpr const combined_string<>
concat() {
  return concat("");
}

live example