我试图创建一个constexpr函数，它将连接Xeo的以下答案来连接任意数量的char数组，它连接两个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都会给出错误，表明找不到与concat表达式中的decltype匹配的函数。
铛：
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.
来自gcc和clang的错误都表明第二个concat函数模板不是concat表达式中decltype的候选者。仅考虑第一个模板。为什么这样，我该如何解决这个问题？
修改：关于为什么decltype无法递归使用的相关问题
trailing return type using decltype with a variadic template function
答案 0 :(得分：13)
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...));
}
摆脱了递归 - decltype。
以下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("");
}
答案 1 :(得分：2)
使用 C++17，解决方案变得非常简单 (here's the live version)：
#include <initializer_list>
// we cannot return a char array from a function, therefore we need a wrapper
template <unsigned N>
struct String {
char c[N];
};
template<unsigned ...Len>
constexpr auto cat(const char (&...strings)[Len]) {
constexpr unsigned N = (... + Len) - sizeof...(Len);
String<N + 1> result = {};
result.c[N] = '\0';
char* dst = result.c;
for (const char* src : {strings...}) {
for (; *src != '\0'; src++, dst++) {
*dst = *src;
}
}
return result;
}
// can be used to build other constexpr functions
template<unsigned L>
constexpr auto makeCopyright(const char (&author)[L]) {
return cat("\xC2\xA9 ", author);
}
constexpr char one[] = "The desert was the apotheosis of all deserts";
constexpr char two[] = "huge, standing to the sky";
constexpr auto three = cat(
cat(one, ", ", two).c, // can concatenate recursively
" ",
"for what looked like eternity in all directions."); // can use in-place literals
constexpr auto phrase = cat(
three.c, // can reuse existing cats
"\n",
makeCopyright("Stephen King").c);
#include <cstdio>
int main() {
puts(phrase.c);
return 0;
}