我对如何在递归链中传递类型信息感到困惑。
以下代码的目标是执行类似python＆＃39; s ",".join(["a","b","c"])的操作，但是递归，以便我可以加入一组数组的数组...的字符串，其中有许多分隔符因为结构很深。
这是我的一些注释（试图让它编译）代码：
#include <iostream>
#include <typeinfo>
#include <vector>
template <typename RAI>
std::string string_join(RAI begin, RAI end, std::string delimiter) {
if (begin == end) { return ""; }
std::string joint = std::to_string(*begin);
begin++;
for (; begin != end; begin++) {
joint += delimiter + std::to_string(*begin);
}
return joint;
}
template <typename RAI, int depth>
struct string_join_recursive {
std::string operator()(RAI iterator, std::string *delimiter, int *dimensions) {
typedef typename std::iterator_traits<RAI>::value_type::iterator value_iterator_type;
std::cout << typeid(value_iterator_type).name() << " " << typeid(std::begin(*iterator)).name() << std::endl;
std::string joint = string_join_recursive<value_iterator_type, depth-1>(std::begin(*iterator), delimiter+1, dimensions+1);
iterator++;
for (int i=1; i<*dimensions; i++) {
//joint += *delimiter + string_join_recursive<value_iterator_type, depth-1>(std::begin(*iterator), delimiter+1, dimensions+1);
iterator++;
}
return "";//joint;
}
};
template <typename RAI>
struct string_join_recursive<RAI,1> {
std::string operator()(RAI iterator, std::string *delimiter, int *dimensions) {
return string_join(iterator, iterator + *dimensions, *delimiter);
}
};
int main() {
std::vector<std::vector<int>> a;
std::vector<int> b = {1,2,3};
std::vector<int> c = {1,2,4};
std::string delimiters[2] = {"\n",","};
int dimensions[2] = {2, 3};
std::cout << string_join_recursive<std::vector<std::vector<int>>::iterator, 2>()(a.begin(), delimiters, dimensions) << std::endl;
}
我没有经验丰富的c ++程序员，所以这可能有不止一个问题，但目前它甚至不会编译。显然value_iterator_type并不是我认为这个程序运行的类型，但我不确定它究竟是什么。 OTOH，当我注释第5行时，它会编译，并且上面的打印行说这两个东西是相同的类型。
这是我得到的错误：
error: no matching function for call to ‘string_join_recursive<__gnu_cxx::__normal_iterator<int*, std::vector<int> >, 1>::string_join_recursive(std::vector<int>::iterator, std::__cxx11::string*, int*)’
std::string joint = string_join_recursive<value_iterator_type, depth-1>(std::begin(*iterator), delimiter+1, dimensions+1);
奖励积分以更清洁的方式做这件事。 有关使用任意迭代（数组等）的方法的更多奖励点
答案 0 :(得分：2)
这是我将如何处理它的草图：
#include <iostream>
#include <algorithm>
#include <type_traits>
#include <sstream>
#include <vector>
#include <tuple>
// this allows us to test if a element should be recursed into
template <typename iterable, typename tag = void>
struct is_iterable
{
static constexpr bool value = false;
};
template <typename iterable>
struct is_iterable <iterable, std::void_t <decltype (std::begin (std::declval <iterable> ()))>>
{
static constexpr bool value = true;
};
template <typename iterable>
constexpr bool is_iterable_v = is_iterable <iterable>::value;
// visit elements of a tuple
template <size_t Index, typename visitor, typename ...types>
void visit_tuple (std::tuple <types...> const & Tuple, visitor && Visitor)
{
Visitor (std::get <Index> (Tuple));
constexpr auto nextIndex = Index + 1;
if constexpr (nextIndex < sizeof... (types))
visit_tuple <nextIndex> (Tuple, Visitor);
}
// this function generically walks the elements of a container, with an overload for tuples and pairs
template <typename container, typename visitor>
std::enable_if_t <is_iterable_v <container>>
visit_elements (container const & Container, visitor && Visitor)
{
for (auto && Element: Container)
Visitor (Element);
}
template <typename visitor, typename ...element_types>
void visit_elements (std::tuple <element_types...> const & Tuple, visitor && Visitor)
{
if constexpr (sizeof... (element_types) > 0)
visit_tuple <0> (Tuple, Visitor);
}
template <typename visitor, typename first_type, typename second_type>
void visit_elements (std::pair <first_type, second_type> const & Pair, visitor && Visitor)
{
Visitor (Pair.first);
Visitor (Pair.second);
}
// type trait for testing if a value is "visitable"
struct dummy_visitor { template <typename type> void operator () (type &&); };
template <typename container, typename tag = void>
struct is_visitable
{
static constexpr bool value = false;
};
template <typename container>
struct is_visitable <container, std::void_t <decltype (visit_elements (std::declval <container> (), dummy_visitor()))>>
{
static constexpr bool value = true;
};
template <typename container>
constexpr bool is_visitable_v = is_visitable <container>::value;
// this function walks each item and either emits it or recurses into it
template <typename iterable>
bool join_impl (std::ostream & os, char delim, bool emitted_anything, iterable const & Iterable)
{
using std::begin;
using std::end;
visit_elements (Iterable, [&] (auto && Element) {
if constexpr (!is_visitable_v <decltype (Element)>)
{
if (emitted_anything)
os << delim;
os << Element;
emitted_anything = true;
}
else
{
emitted_anything = join_impl (os, delim, emitted_anything, Element);
}
});
return emitted_anything;
}
// these are wrappers to adapt join_impl for different use cases
template <typename container>
struct joiner
{
char delim;
container const & Container;
operator std::string () const { return to_string <char> (); }
template <typename char_t>
std::basic_string <char_t> to_string () const;
};
template <typename container>
std::ostream & operator << (std::ostream & os, joiner <container> j)
{
bool emitted_anything = false;
join_impl (os, j.delim, emitted_anything, j.Container);
return os;
}
template <typename container>
template <typename char_t>
std::basic_string <char_t> joiner <container>::to_string () const
{
std::ostringstream os;
os << *this;
return os;
}
template <typename container>
std::enable_if_t <is_visitable_v <container>, joiner <container>>
join (char delim, container const & Container)
{
return joiner <container> { delim, Container };
}
// test the streaming use case
int main ()
{
std::vector <std::vector <std::tuple <int, float>>> x {
{ { 1, 1.0f }, { 2, 2.0f }},
{ { 3, 3.0f }, { 4, 4.0f }},
};
std::cout << join (',', x) << std::endl;
}
这应该只适用于支持标准迭代协议的任何递归。它还支持元组和对。 In可以通过定义访问每个暴露字段的visit_elements重载来扩展到用户定义结构。 is_iterable类型特征需要在其预期的界面中更精确地工作。
答案 1 :(得分：0)
我设法清理错误。我只是在误导一个仿函数。我最初的意图是使它适用于数组，但事实并非如此。无论如何，这对于任意维度的csv来说都是一个csv-writer。
#include <iostream>
#include <typeinfo>
#include <vector>
template <typename RAI>
std::string string_join(RAI begin, RAI end, std::string delimiter) {
if (begin == end) { return ""; }
std::string joint = std::to_string(*begin);
begin++;
for (; begin != end; begin++) {
joint += delimiter + std::to_string(*begin);
}
return joint;
}
template <typename RAI, int depth>
struct string_join_recursive {
std::string operator()(RAI iterator, std::string *delimiters, int *dimensions) {
typedef typename std::iterator_traits<RAI>::value_type::iterator value_iterator_type;
std::string joint = string_join_recursive<value_iterator_type, depth-1>()(std::begin(*iterator), delimiters+1, dimensions+1);
iterator++;
for (int i=1; i<*dimensions; i++) {
joint += *delimiters + string_join_recursive<value_iterator_type, depth-1>()(std::begin(*iterator), delimiters+1, dimensions+1);
iterator++;
}
return joint;
}
};
template <typename RAI>
struct string_join_recursive<RAI,1> {
std::string operator()(RAI iterator, std::string *delimiters, int *dimensions) {
return string_join(iterator, iterator + *dimensions, *delimiters);
}
};
int main() {
std::vector<int> a = {1,2,3};
std::vector<int> b = {4,5,6};
std::vector<int> c = {7,8,9};
std::vector<int> d = {10,11,12};
std::vector<std::vector<int>> e = {a,b};
std::vector<std::vector<int>> f = {c,d};
std::vector<std::vector<std::vector<int>>> g = {e,f};
std::string delimiters[3] = {"\n",";",","};
int dimensions[3] = {2, 2, 3};
std::cout << string_join_recursive<std::vector<std::vector<std::vector<int>>>::iterator, 3>()(g.begin(), delimiters, dimensions) << std::endl;
}