我可以将 BOOST_FUSION_ADAPT_STRUCT 用于继承的东西吗?
Can I use BOOST_FUSION_ADAPT_STRUCT with inherited stuff?
假设我有
struct cat
{
int tail;
int head;
};
struct bird
{
int wing;
int bursa;
};
如果我这样做...
struct wat : public cat, public bird
{
};
BOOST_FUSION_ADAPT_STRUCT(cat,tail,head)
BOOST_FUSION_ADAPT_STRUCT(bird, wing, bursa)
BOOST_FUSION_ADAPT_STRUCT(wat, wat::cat, wat::bird)
...我无法构建,但如果我像下面这样显式引用继承的对象,它是完全有效的。
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
namespace qi = boost::spirit::qi;
struct wat
{
public:
cat z;
bird q;
};
BOOST_FUSION_ADAPT_STRUCT(cat,tail,head)
BOOST_FUSION_ADAPT_STRUCT(bird, wing, bursa)
BOOST_FUSION_ADAPT_STRUCT(wat, z, q)
有什么方法可以让第一个版本正常工作,这样我就可以调整继承的 public 成员的结构吗?我绝对不想做 BOOST_FUSION_ADAPT_STRUCT(wat,tail,head,wing,bursa)
但这似乎是我能找到的继承成员到达那里的唯一方法。
这里有类似的问题:c++/boost fusion handle parent class
简短的问题:不,这不是一项功能。
您可以使用聚合而不是继承,
struct wat {
cat _cat;
bird _bird;
};
BOOST_FUSION_ADAPT_STRUCT(wat, _cat, _bird)
但你不会神奇地得到一个扁平的序列。
您可能想在处理融合序列的代码中做的是编写对已知碱基 类 的支持的代码,并在适应的序列成员之外处理它们。
带聚合的演示
#include <boost/fusion/include/adapted.hpp>
struct cat {
int tail;
int head;
};
struct bird {
int wing;
int cloaca;
};
struct wat {
cat _cat;
bird _bird;
};
BOOST_FUSION_ADAPT_STRUCT(cat, tail, head)
BOOST_FUSION_ADAPT_STRUCT(bird, wing, cloaca)
BOOST_FUSION_ADAPT_STRUCT(wat, _cat, _bird)
#include <iostream>
#include <boost/fusion/include/at_c.hpp>
template <typename T, int N = 0> void print(T const& obj) {
namespace fus = boost::fusion;
if constexpr (fus::traits::is_sequence<T>::value) {
if (N==0)
std::cout << "{";
if constexpr (N < fus::size(obj).value) {
auto name = boost::fusion::extension::struct_member_name<T, N>::call();
std::cout << ' ' << name << '=';
print(fus::at_c<N>(obj));
std::cout << ';';
print<T, N+1>(obj);
} else {
std::cout << " }";
}
} else {
std::cout << obj;
}
}
int main() {
print(wat { {1,2}, {3,4} });
}
版画
{ _cat={ tail=1; head=2; }; _bird={ wing=3; cloaca=4; }; }
带有硬编码碱基列表的演示
#include <boost/fusion/include/adapted.hpp>
struct cat {
int tail;
int head;
};
struct bird {
int wing;
int cloaca;
};
struct wat : cat, bird {
int something;
int extra;
wat(int tail, int head, int wing, int cloaca, int something, int extra)
: cat{tail, head}, bird{wing, cloaca}, something(something), extra(extra)
{ }
};
BOOST_FUSION_ADAPT_STRUCT(cat, tail, head)
BOOST_FUSION_ADAPT_STRUCT(bird, wing, cloaca)
BOOST_FUSION_ADAPT_STRUCT(wat, something, extra)
#include <iostream>
#include <boost/fusion/include/at_c.hpp>
template <typename... KnownBases>
struct Demo {
template <typename T, int N = 0> static void print(T const& obj, bool outer_sequence_braces = true) {
namespace fus = boost::fusion;
if constexpr (fus::traits::is_sequence<T>::value) {
if (N==0)
{
if (outer_sequence_braces) std::cout << "{";
print_bases<KnownBases...>(obj);
}
if constexpr (N < fus::size(obj).value) {
auto name = boost::fusion::extension::struct_member_name<T, N>::call();
std::cout << ' ' << name << '=';
print(fus::at_c<N>(obj), true/*inner sequences get braces*/);
std::cout << ';';
print<T, N+1>(obj, outer_sequence_braces);
} else {
if (outer_sequence_braces) std::cout << " }";
}
} else {
std::cout << obj;
}
}
template <typename Base, typename T> static bool print_base(T const& obj) {
if constexpr (not std::is_same<Base, T>() && std::is_base_of<Base, T>())
print(static_cast<Base const&>(obj), false);
return true;
}
template <typename... Bases, typename T> static void print_bases(T const& obj) {
bool discard[] = { print_base<Bases>(obj)... };
(void) discard;
}
};
int main() {
Demo<cat, bird>::print(wat { 1, 2, 3, 4, 5, 6 });
}
版画
{ tail=1; head=2; wing=3; cloaca=4; something=5; extra=6; }
假设我有
struct cat
{
int tail;
int head;
};
struct bird
{
int wing;
int bursa;
};
如果我这样做...
struct wat : public cat, public bird
{
};
BOOST_FUSION_ADAPT_STRUCT(cat,tail,head)
BOOST_FUSION_ADAPT_STRUCT(bird, wing, bursa)
BOOST_FUSION_ADAPT_STRUCT(wat, wat::cat, wat::bird)
...我无法构建,但如果我像下面这样显式引用继承的对象,它是完全有效的。
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
namespace qi = boost::spirit::qi;
struct wat
{
public:
cat z;
bird q;
};
BOOST_FUSION_ADAPT_STRUCT(cat,tail,head)
BOOST_FUSION_ADAPT_STRUCT(bird, wing, bursa)
BOOST_FUSION_ADAPT_STRUCT(wat, z, q)
有什么方法可以让第一个版本正常工作,这样我就可以调整继承的 public 成员的结构吗?我绝对不想做 BOOST_FUSION_ADAPT_STRUCT(wat,tail,head,wing,bursa)
但这似乎是我能找到的继承成员到达那里的唯一方法。
这里有类似的问题:c++/boost fusion handle parent class
简短的问题:不,这不是一项功能。
您可以使用聚合而不是继承,
struct wat {
cat _cat;
bird _bird;
};
BOOST_FUSION_ADAPT_STRUCT(wat, _cat, _bird)
但你不会神奇地得到一个扁平的序列。
您可能想在处理融合序列的代码中做的是编写对已知碱基 类 的支持的代码,并在适应的序列成员之外处理它们。
带聚合的演示
#include <boost/fusion/include/adapted.hpp>
struct cat {
int tail;
int head;
};
struct bird {
int wing;
int cloaca;
};
struct wat {
cat _cat;
bird _bird;
};
BOOST_FUSION_ADAPT_STRUCT(cat, tail, head)
BOOST_FUSION_ADAPT_STRUCT(bird, wing, cloaca)
BOOST_FUSION_ADAPT_STRUCT(wat, _cat, _bird)
#include <iostream>
#include <boost/fusion/include/at_c.hpp>
template <typename T, int N = 0> void print(T const& obj) {
namespace fus = boost::fusion;
if constexpr (fus::traits::is_sequence<T>::value) {
if (N==0)
std::cout << "{";
if constexpr (N < fus::size(obj).value) {
auto name = boost::fusion::extension::struct_member_name<T, N>::call();
std::cout << ' ' << name << '=';
print(fus::at_c<N>(obj));
std::cout << ';';
print<T, N+1>(obj);
} else {
std::cout << " }";
}
} else {
std::cout << obj;
}
}
int main() {
print(wat { {1,2}, {3,4} });
}
版画
{ _cat={ tail=1; head=2; }; _bird={ wing=3; cloaca=4; }; }
带有硬编码碱基列表的演示
#include <boost/fusion/include/adapted.hpp>
struct cat {
int tail;
int head;
};
struct bird {
int wing;
int cloaca;
};
struct wat : cat, bird {
int something;
int extra;
wat(int tail, int head, int wing, int cloaca, int something, int extra)
: cat{tail, head}, bird{wing, cloaca}, something(something), extra(extra)
{ }
};
BOOST_FUSION_ADAPT_STRUCT(cat, tail, head)
BOOST_FUSION_ADAPT_STRUCT(bird, wing, cloaca)
BOOST_FUSION_ADAPT_STRUCT(wat, something, extra)
#include <iostream>
#include <boost/fusion/include/at_c.hpp>
template <typename... KnownBases>
struct Demo {
template <typename T, int N = 0> static void print(T const& obj, bool outer_sequence_braces = true) {
namespace fus = boost::fusion;
if constexpr (fus::traits::is_sequence<T>::value) {
if (N==0)
{
if (outer_sequence_braces) std::cout << "{";
print_bases<KnownBases...>(obj);
}
if constexpr (N < fus::size(obj).value) {
auto name = boost::fusion::extension::struct_member_name<T, N>::call();
std::cout << ' ' << name << '=';
print(fus::at_c<N>(obj), true/*inner sequences get braces*/);
std::cout << ';';
print<T, N+1>(obj, outer_sequence_braces);
} else {
if (outer_sequence_braces) std::cout << " }";
}
} else {
std::cout << obj;
}
}
template <typename Base, typename T> static bool print_base(T const& obj) {
if constexpr (not std::is_same<Base, T>() && std::is_base_of<Base, T>())
print(static_cast<Base const&>(obj), false);
return true;
}
template <typename... Bases, typename T> static void print_bases(T const& obj) {
bool discard[] = { print_base<Bases>(obj)... };
(void) discard;
}
};
int main() {
Demo<cat, bird>::print(wat { 1, 2, 3, 4, 5, 6 });
}
版画
{ tail=1; head=2; wing=3; cloaca=4; something=5; extra=6; }