将 Boost Spirit 解析器从 boost::variant 过渡到 std::variant
Transitioning Boost Spirit parser from boost::variant to std::variant
我目前正在尝试将一些代码从使用 boost::variant 移到 std::variant,但是 运行 变成了一个我无法弄清楚的问题。下面是一个最小的测试用例:
#include <string>
#include <variant>
#include <boost/spirit/home/x3.hpp>
#include <boost/variant/recursive_wrapper.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
struct Recurse;
//using Base = boost::variant< // This works
using Base = std::variant<
std::string,
boost::recursive_wrapper<Recurse>>;
struct Recurse
{
int _i;
Base _base = std::string{};
};
BOOST_FUSION_ADAPT_STRUCT(
Recurse,
(int, _i),
(Base, _base)
)
namespace x3 = boost::spirit::x3;
namespace ascii = boost::spirit::x3::ascii;
const x3::rule<class Base_, Base> base = "base";
const auto operand = *x3::char_("a-zA-Z0-9_") | base;
const auto base_def = (x3::int_ >> operand) | operand;
BOOST_SPIRIT_DEFINE(base)
int main()
{
std::string text;
Base result;
x3::phrase_parse(std::begin(text), std::end(text), base, ascii::space, result);
return 0;
}
我认为 正在发生的是解析器试图将 int 直接分配给类型 Base
的值,但由于 int 不直接映射到 std::string 或 boost::recursive_wrapper<>,它会变得不高兴(因此不高兴我的意思是 11 页的编译器错误)。 以某种方式、boost::variant 避免了这个问题。请问有什么线索吗?
Somehow boost::variant
avoids the error.
是的。 Boost变体具有属性传播
支持。
此外,boost::variant
对 boost::recursive_wrapper
进行了特殊处理,因此它可能是双重 no-fly。
A good article about recursive std::variant
s is here https://vittorioromeo.info/index/blog/variants_lambdas_part_2.html
boost::variant
怎么了?
如果您愿意,可以写一些转换特征,甚至可以查看 x3::variant - 它可能更适合您?
#include <string>
#include <boost/spirit/home/x3.hpp>
#include <boost/spirit/home/x3/support/ast/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
namespace x3 = boost::spirit::x3;
namespace ascii = boost::spirit::x3::ascii;
struct Recurse;
using Base = x3::variant<
std::string,
x3::forward_ast<Recurse> >;
struct Recurse
{
int _i;
Base _base;
};
BOOST_FUSION_ADAPT_STRUCT(
Recurse,
(int, _i),
(Base, _base)
)
const x3::rule<class Base_, Base> base = "base";
const auto operand = *x3::char_("a-zA-Z0-9_") | base;
const auto base_def = (x3::int_ >> operand) | operand;
BOOST_SPIRIT_DEFINE(base)
int main()
{
std::string text;
Base result;
x3::phrase_parse(std::begin(text), std::end(text), base, ascii::space, result);
return 0;
}
Side note: No x3::forward_ast<>
does not help with std::variant
, confirming that std::variant
just lacks support in x3
更新
您可以 work-around 通过使您的 Base
具有所需机制的派生结构来向 Spirit 表明它是一个变体(以及在哪些类型上)。这样你就不必经历特质专业化地狱:
struct Recurse;
struct Base : std::variant<std::string, boost::recursive_wrapper<Recurse> > {
using BaseV = std::variant<std::string, boost::recursive_wrapper<Recurse> >;
using BaseV::BaseV;
using BaseV::operator=;
struct adapted_variant_tag {};
using types = boost::mpl::list<std::string, Recurse>;
};
struct Recurse {
int _i;
Base _base;
};
如您所见,它基本相同¹,但添加了 adapted_variant_tag
和 types
嵌套类型。
注意 通过巧妙地对 types
序列进行硬编码,我们可以假装巧妙地处理递归包装器。我们很幸运,这足以欺骗系统。
添加一些调试输出和 test-cases:
#include <string>
#include <variant>
#include <iostream>
#include <iomanip>
#include <boost/spirit/home/x3.hpp>
#include <boost/spirit/home/x3/support/ast/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
namespace x3 = boost::spirit::x3;
namespace ascii = boost::spirit::x3::ascii;
namespace { // for debug
template<class T>
std::ostream& operator<<(std::ostream& os, boost::recursive_wrapper<T> const& rw) {
return os << rw.get();
}
template<class... Ts>
std::ostream& operator<<(std::ostream& os, std::variant<Ts...> const& sv) {
std::visit([&os](const auto& v) { os << v; }, sv);
return os;
}
}
struct Recurse;
struct Base : std::variant<std::string, boost::recursive_wrapper<Recurse> > {
using BaseV = std::variant<std::string, boost::recursive_wrapper<Recurse> >;
using BaseV::BaseV;
using BaseV::operator=;
struct adapted_variant_tag {};
using types = boost::mpl::list<std::string, Recurse>;
};
struct Recurse {
int _i;
Base _base;
friend std::ostream& operator<<(std::ostream& os, Recurse const& r) {
return os << "[" << r._i << ", " << r._base << "]";
}
};
BOOST_FUSION_ADAPT_STRUCT(
Recurse,
(int, _i),
(Base, _base)
)
static_assert(x3::traits::is_variant<Base>::value);
const x3::rule<class Base_, Base> base = "base";
const auto operand = *x3::char_("a-zA-Z0-9_") | base;
const auto base_def = (x3::int_ >> operand) | operand;
BOOST_SPIRIT_DEFINE(base)
int main()
{
for (std::string const text : { "yeah8", "32 more" }) {
Base result;
auto f = begin(text), l = end(text);
if (x3::phrase_parse(f, l, base, ascii::space, result)) {
std::cout << "Result: " << result << "\n";
} else {
std::cout << "Failed\n";
}
if (f!=l) {
std::cout << "Remaining input: " << std::quoted(std::string(f,l)) << "\n";
}
}
}
打印
Result: yeah8
Result: [32, more]
更新 2:锦上添花
以下是使 std::variant
正常工作所需的特征:
namespace boost::spirit::x3::traits {
template<typename... t>
struct is_variant<std::variant<t...> >
: mpl::true_ {};
template <typename attribute, typename... t>
struct variant_has_substitute_impl<std::variant<t...>, attribute>
{
typedef std::variant<t...> variant_type;
typedef typename mpl::transform<
mpl::list<t...>
, unwrap_recursive<mpl::_1>
>::type types;
typedef typename mpl::end<types>::type end;
typedef typename mpl::find<types, attribute>::type iter_1;
typedef typename
mpl::eval_if<
is_same<iter_1, end>,
mpl::find_if<types, traits::is_substitute<mpl::_1, attribute>>,
mpl::identity<iter_1>
>::type
iter;
typedef mpl::not_<is_same<iter, end>> type;
};
template <typename attribute, typename... t>
struct variant_find_substitute<std::variant<t...>, attribute>
{
typedef std::variant<t...> variant_type;
typedef typename mpl::transform<
mpl::list<t...>
, unwrap_recursive<mpl::_1>
>::type types;
typedef typename mpl::end<types>::type end;
typedef typename mpl::find<types, attribute>::type iter_1;
typedef typename
mpl::eval_if<
is_same<iter_1, end>,
mpl::find_if<types, traits::is_substitute<mpl::_1, attribute> >,
mpl::identity<iter_1>
>::type
iter;
typedef typename
mpl::eval_if<
is_same<iter, end>,
mpl::identity<attribute>,
mpl::deref<iter>
>::type
type;
};
template <typename... t>
struct variant_find_substitute<std::variant<t...>, std::variant<t...> >
: mpl::identity<std::variant<t...> > {};
}
噪音很大,但你可以把它放在 header 的某个地方。
奖金
修复语法:
- 您可能打算
lexeme[]
围绕字符串生成
- 您可能想要字符串的最小长度(+char_,而不是 *char_),因为没有分隔符
- 您可能必须重新排序分支,因为字符串生成会吞噬递归规则的整数。
这是我的 touched-up 语法,其中的规则与 AST 非常相似,通常是有道理的:
namespace Parser {
static_assert(x3::traits::is_variant<Base>::value);
const x3::rule<class Base_, Base> base = "base";
const auto string = x3::lexeme[+x3::char_("a-zA-Z0-9_")];
const auto recurse = x3::int_ >> base;
const auto base_def = recurse | string;
BOOST_SPIRIT_DEFINE(base)
}
简化融合
最后但同样重要的是,在C++11时代你可以推导出适应的融合成员:
BOOST_FUSION_ADAPT_STRUCT(Recurse, _i, _base)
现场完整演示
#include <string>
#include <variant>
#include <iostream>
#include <iomanip>
#include <boost/spirit/home/x3.hpp>
#include <boost/spirit/home/x3/support/ast/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
namespace x3 = boost::spirit::x3;
namespace ascii = boost::spirit::x3::ascii;
namespace { // for debug
template<class T>
std::ostream& operator<<(std::ostream& os, boost::recursive_wrapper<T> const& rw) {
return os << rw.get();
}
template<class... Ts>
std::ostream& operator<<(std::ostream& os, std::variant<Ts...> const& sv) {
std::visit([&os](const auto& v) { os << v; }, sv);
return os;
}
}
struct Recurse;
using Base = std::variant<
std::string,
boost::recursive_wrapper<Recurse> >;
namespace boost::spirit::x3::traits {
template<typename... T>
struct is_variant<std::variant<T...> >
: mpl::true_ {};
template <typename Attribute, typename... T>
struct variant_has_substitute_impl<std::variant<T...>, Attribute>
{
typedef std::variant<T...> variant_type;
typedef typename mpl::transform<
mpl::list<T...>
, unwrap_recursive<mpl::_1>
>::type types;
typedef typename mpl::end<types>::type end;
typedef typename mpl::find<types, Attribute>::type iter_1;
typedef typename
mpl::eval_if<
is_same<iter_1, end>,
mpl::find_if<types, traits::is_substitute<mpl::_1, Attribute>>,
mpl::identity<iter_1>
>::type
iter;
typedef mpl::not_<is_same<iter, end>> type;
};
template <typename Attribute, typename... T>
struct variant_find_substitute<std::variant<T...>, Attribute>
{
typedef std::variant<T...> variant_type;
typedef typename mpl::transform<
mpl::list<T...>
, unwrap_recursive<mpl::_1>
>::type types;
typedef typename mpl::end<types>::type end;
typedef typename mpl::find<types, Attribute>::type iter_1;
typedef typename
mpl::eval_if<
is_same<iter_1, end>,
mpl::find_if<types, traits::is_substitute<mpl::_1, Attribute> >,
mpl::identity<iter_1>
>::type
iter;
typedef typename
mpl::eval_if<
is_same<iter, end>,
mpl::identity<Attribute>,
mpl::deref<iter>
>::type
type;
};
template <typename... T>
struct variant_find_substitute<std::variant<T...>, std::variant<T...> >
: mpl::identity<std::variant<T...> > {};
}
static_assert(x3::traits::is_variant<Base>{}, "");
struct Recurse
{
int _i;
Base _base;
friend std::ostream& operator<<(std::ostream& os, Recurse const& r) {
return os << "[" << r._i << ", " << r._base << "]";
}
};
BOOST_FUSION_ADAPT_STRUCT(Recurse, _i, _base)
namespace Parser {
static_assert(x3::traits::is_variant<Base>::value);
const x3::rule<class Base_, Base> base = "base";
const auto string = x3::lexeme[+x3::char_("a-zA-Z0-9_")];
const auto recurse = x3::int_ >> base;
const auto base_def = recurse | string;
BOOST_SPIRIT_DEFINE(base)
}
int main()
{
for (std::string const text : { "yeah8", "32 more", "18 766 most" }) {
Base result;
auto f = begin(text), l = end(text);
if (x3::phrase_parse(f, l, Parser::base, ascii::space, result)) {
std::cout << "Result: " << result << "\n";
} else {
std::cout << "Failed\n";
}
if (f!=l) {
std::cout << "Remaining input: " << std::quoted(std::string(f,l)) << "\n";
}
}
}
打印:
Result: yeah8
Result: [32, more]
Result: [18, [766, most]]
¹(在需要显式访问 base-class 的泛型编程中,细微差别可能会影响您)
我目前正在尝试将一些代码从使用 boost::variant 移到 std::variant,但是 运行 变成了一个我无法弄清楚的问题。下面是一个最小的测试用例:
#include <string>
#include <variant>
#include <boost/spirit/home/x3.hpp>
#include <boost/variant/recursive_wrapper.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
struct Recurse;
//using Base = boost::variant< // This works
using Base = std::variant<
std::string,
boost::recursive_wrapper<Recurse>>;
struct Recurse
{
int _i;
Base _base = std::string{};
};
BOOST_FUSION_ADAPT_STRUCT(
Recurse,
(int, _i),
(Base, _base)
)
namespace x3 = boost::spirit::x3;
namespace ascii = boost::spirit::x3::ascii;
const x3::rule<class Base_, Base> base = "base";
const auto operand = *x3::char_("a-zA-Z0-9_") | base;
const auto base_def = (x3::int_ >> operand) | operand;
BOOST_SPIRIT_DEFINE(base)
int main()
{
std::string text;
Base result;
x3::phrase_parse(std::begin(text), std::end(text), base, ascii::space, result);
return 0;
}
我认为 正在发生的是解析器试图将 int 直接分配给类型 Base
的值,但由于 int 不直接映射到 std::string 或 boost::recursive_wrapper<>,它会变得不高兴(因此不高兴我的意思是 11 页的编译器错误)。 以某种方式、boost::variant 避免了这个问题。请问有什么线索吗?
Somehow
boost::variant
avoids the error.
是的。 Boost变体具有属性传播 支持。
此外,boost::variant
对 boost::recursive_wrapper
进行了特殊处理,因此它可能是双重 no-fly。
A good article about recursive
std::variant
s is here https://vittorioromeo.info/index/blog/variants_lambdas_part_2.html
boost::variant
怎么了?
如果您愿意,可以写一些转换特征,甚至可以查看 x3::variant - 它可能更适合您?
#include <string>
#include <boost/spirit/home/x3.hpp>
#include <boost/spirit/home/x3/support/ast/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
namespace x3 = boost::spirit::x3;
namespace ascii = boost::spirit::x3::ascii;
struct Recurse;
using Base = x3::variant<
std::string,
x3::forward_ast<Recurse> >;
struct Recurse
{
int _i;
Base _base;
};
BOOST_FUSION_ADAPT_STRUCT(
Recurse,
(int, _i),
(Base, _base)
)
const x3::rule<class Base_, Base> base = "base";
const auto operand = *x3::char_("a-zA-Z0-9_") | base;
const auto base_def = (x3::int_ >> operand) | operand;
BOOST_SPIRIT_DEFINE(base)
int main()
{
std::string text;
Base result;
x3::phrase_parse(std::begin(text), std::end(text), base, ascii::space, result);
return 0;
}
Side note: No
x3::forward_ast<>
does not help withstd::variant
, confirming thatstd::variant
just lacks support in x3
更新
您可以 work-around 通过使您的 Base
具有所需机制的派生结构来向 Spirit 表明它是一个变体(以及在哪些类型上)。这样你就不必经历特质专业化地狱:
struct Recurse;
struct Base : std::variant<std::string, boost::recursive_wrapper<Recurse> > {
using BaseV = std::variant<std::string, boost::recursive_wrapper<Recurse> >;
using BaseV::BaseV;
using BaseV::operator=;
struct adapted_variant_tag {};
using types = boost::mpl::list<std::string, Recurse>;
};
struct Recurse {
int _i;
Base _base;
};
如您所见,它基本相同¹,但添加了 adapted_variant_tag
和 types
嵌套类型。
注意 通过巧妙地对 types
序列进行硬编码,我们可以假装巧妙地处理递归包装器。我们很幸运,这足以欺骗系统。
添加一些调试输出和 test-cases:
#include <string>
#include <variant>
#include <iostream>
#include <iomanip>
#include <boost/spirit/home/x3.hpp>
#include <boost/spirit/home/x3/support/ast/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
namespace x3 = boost::spirit::x3;
namespace ascii = boost::spirit::x3::ascii;
namespace { // for debug
template<class T>
std::ostream& operator<<(std::ostream& os, boost::recursive_wrapper<T> const& rw) {
return os << rw.get();
}
template<class... Ts>
std::ostream& operator<<(std::ostream& os, std::variant<Ts...> const& sv) {
std::visit([&os](const auto& v) { os << v; }, sv);
return os;
}
}
struct Recurse;
struct Base : std::variant<std::string, boost::recursive_wrapper<Recurse> > {
using BaseV = std::variant<std::string, boost::recursive_wrapper<Recurse> >;
using BaseV::BaseV;
using BaseV::operator=;
struct adapted_variant_tag {};
using types = boost::mpl::list<std::string, Recurse>;
};
struct Recurse {
int _i;
Base _base;
friend std::ostream& operator<<(std::ostream& os, Recurse const& r) {
return os << "[" << r._i << ", " << r._base << "]";
}
};
BOOST_FUSION_ADAPT_STRUCT(
Recurse,
(int, _i),
(Base, _base)
)
static_assert(x3::traits::is_variant<Base>::value);
const x3::rule<class Base_, Base> base = "base";
const auto operand = *x3::char_("a-zA-Z0-9_") | base;
const auto base_def = (x3::int_ >> operand) | operand;
BOOST_SPIRIT_DEFINE(base)
int main()
{
for (std::string const text : { "yeah8", "32 more" }) {
Base result;
auto f = begin(text), l = end(text);
if (x3::phrase_parse(f, l, base, ascii::space, result)) {
std::cout << "Result: " << result << "\n";
} else {
std::cout << "Failed\n";
}
if (f!=l) {
std::cout << "Remaining input: " << std::quoted(std::string(f,l)) << "\n";
}
}
}
打印
Result: yeah8
Result: [32, more]
更新 2:锦上添花
以下是使 std::variant
正常工作所需的特征:
namespace boost::spirit::x3::traits {
template<typename... t>
struct is_variant<std::variant<t...> >
: mpl::true_ {};
template <typename attribute, typename... t>
struct variant_has_substitute_impl<std::variant<t...>, attribute>
{
typedef std::variant<t...> variant_type;
typedef typename mpl::transform<
mpl::list<t...>
, unwrap_recursive<mpl::_1>
>::type types;
typedef typename mpl::end<types>::type end;
typedef typename mpl::find<types, attribute>::type iter_1;
typedef typename
mpl::eval_if<
is_same<iter_1, end>,
mpl::find_if<types, traits::is_substitute<mpl::_1, attribute>>,
mpl::identity<iter_1>
>::type
iter;
typedef mpl::not_<is_same<iter, end>> type;
};
template <typename attribute, typename... t>
struct variant_find_substitute<std::variant<t...>, attribute>
{
typedef std::variant<t...> variant_type;
typedef typename mpl::transform<
mpl::list<t...>
, unwrap_recursive<mpl::_1>
>::type types;
typedef typename mpl::end<types>::type end;
typedef typename mpl::find<types, attribute>::type iter_1;
typedef typename
mpl::eval_if<
is_same<iter_1, end>,
mpl::find_if<types, traits::is_substitute<mpl::_1, attribute> >,
mpl::identity<iter_1>
>::type
iter;
typedef typename
mpl::eval_if<
is_same<iter, end>,
mpl::identity<attribute>,
mpl::deref<iter>
>::type
type;
};
template <typename... t>
struct variant_find_substitute<std::variant<t...>, std::variant<t...> >
: mpl::identity<std::variant<t...> > {};
}
噪音很大,但你可以把它放在 header 的某个地方。
奖金
修复语法:
- 您可能打算
lexeme[]
围绕字符串生成 - 您可能想要字符串的最小长度(+char_,而不是 *char_),因为没有分隔符
- 您可能必须重新排序分支,因为字符串生成会吞噬递归规则的整数。
这是我的 touched-up 语法,其中的规则与 AST 非常相似,通常是有道理的:
namespace Parser {
static_assert(x3::traits::is_variant<Base>::value);
const x3::rule<class Base_, Base> base = "base";
const auto string = x3::lexeme[+x3::char_("a-zA-Z0-9_")];
const auto recurse = x3::int_ >> base;
const auto base_def = recurse | string;
BOOST_SPIRIT_DEFINE(base)
}
简化融合
最后但同样重要的是,在C++11时代你可以推导出适应的融合成员:
BOOST_FUSION_ADAPT_STRUCT(Recurse, _i, _base)
现场完整演示
#include <string>
#include <variant>
#include <iostream>
#include <iomanip>
#include <boost/spirit/home/x3.hpp>
#include <boost/spirit/home/x3/support/ast/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
namespace x3 = boost::spirit::x3;
namespace ascii = boost::spirit::x3::ascii;
namespace { // for debug
template<class T>
std::ostream& operator<<(std::ostream& os, boost::recursive_wrapper<T> const& rw) {
return os << rw.get();
}
template<class... Ts>
std::ostream& operator<<(std::ostream& os, std::variant<Ts...> const& sv) {
std::visit([&os](const auto& v) { os << v; }, sv);
return os;
}
}
struct Recurse;
using Base = std::variant<
std::string,
boost::recursive_wrapper<Recurse> >;
namespace boost::spirit::x3::traits {
template<typename... T>
struct is_variant<std::variant<T...> >
: mpl::true_ {};
template <typename Attribute, typename... T>
struct variant_has_substitute_impl<std::variant<T...>, Attribute>
{
typedef std::variant<T...> variant_type;
typedef typename mpl::transform<
mpl::list<T...>
, unwrap_recursive<mpl::_1>
>::type types;
typedef typename mpl::end<types>::type end;
typedef typename mpl::find<types, Attribute>::type iter_1;
typedef typename
mpl::eval_if<
is_same<iter_1, end>,
mpl::find_if<types, traits::is_substitute<mpl::_1, Attribute>>,
mpl::identity<iter_1>
>::type
iter;
typedef mpl::not_<is_same<iter, end>> type;
};
template <typename Attribute, typename... T>
struct variant_find_substitute<std::variant<T...>, Attribute>
{
typedef std::variant<T...> variant_type;
typedef typename mpl::transform<
mpl::list<T...>
, unwrap_recursive<mpl::_1>
>::type types;
typedef typename mpl::end<types>::type end;
typedef typename mpl::find<types, Attribute>::type iter_1;
typedef typename
mpl::eval_if<
is_same<iter_1, end>,
mpl::find_if<types, traits::is_substitute<mpl::_1, Attribute> >,
mpl::identity<iter_1>
>::type
iter;
typedef typename
mpl::eval_if<
is_same<iter, end>,
mpl::identity<Attribute>,
mpl::deref<iter>
>::type
type;
};
template <typename... T>
struct variant_find_substitute<std::variant<T...>, std::variant<T...> >
: mpl::identity<std::variant<T...> > {};
}
static_assert(x3::traits::is_variant<Base>{}, "");
struct Recurse
{
int _i;
Base _base;
friend std::ostream& operator<<(std::ostream& os, Recurse const& r) {
return os << "[" << r._i << ", " << r._base << "]";
}
};
BOOST_FUSION_ADAPT_STRUCT(Recurse, _i, _base)
namespace Parser {
static_assert(x3::traits::is_variant<Base>::value);
const x3::rule<class Base_, Base> base = "base";
const auto string = x3::lexeme[+x3::char_("a-zA-Z0-9_")];
const auto recurse = x3::int_ >> base;
const auto base_def = recurse | string;
BOOST_SPIRIT_DEFINE(base)
}
int main()
{
for (std::string const text : { "yeah8", "32 more", "18 766 most" }) {
Base result;
auto f = begin(text), l = end(text);
if (x3::phrase_parse(f, l, Parser::base, ascii::space, result)) {
std::cout << "Result: " << result << "\n";
} else {
std::cout << "Failed\n";
}
if (f!=l) {
std::cout << "Remaining input: " << std::quoted(std::string(f,l)) << "\n";
}
}
}
打印:
Result: yeah8
Result: [32, more]
Result: [18, [766, most]]
¹(在需要显式访问 base-class 的泛型编程中,细微差别可能会影响您)