模拟基于范围的循环的开始/结束行为 [英] Simulating the range-based for loop's begin/end behavior
问题描述
考虑基于范围的循环的 begin-expr 和 end-expr (N4140 [stmt.ranged] / p1)的规范。给定范围 __范围
类型 _RangeT
,
begin-expr 和 end-expr 的确定方式如下:
$ b $如果
_RangeT
是数组类型, begin-expr 和 end-expr 是<$ c $ b- c> __ range 和
__ range + __bound
,其中__ bound
是
数组绑定。如果_RangeT
是未知大小的数组或
不完整类型的数组,则程序格式不正确;
- 如果
_RangeT
是一个类类型,那么 unqualified-idbegin
和end
在class_RangeT
的范围内查找,就像通过类成员访问
lookup(3.4.5)一样, (或两者)至少发现一个
声明,则 begin-expr 和 end-expr 为__ range.begin()
和
__ range.end()
;
- 否则, begin -expr 和 end-expr 是
begin(__ range)
和end(__ range)
开始和end
分别在
和相关的命名空间(3.4。 2)。 [注意:不执行普通的不合格的
查找(3.4.1)。 - end note ]
是否可以模拟这个 magic_begin 和一个 magic_end
函数模板,使得
for(auto&& p:range_init){/ * statements * /}
和
{
auto&& my_range = range_init;
for(auto b = magic_begin(my_range),e = magic_end(my_range); b!= e; ++ b){
auto&& p = * b;
/ *语句* /
}
}
这个行为完全相同吗?
非答案包括对 std :: begin
/ std :: end
(不处理第三个项目符号等等)和使用std :: begin;开始(范围);
因为,如果ADL for begin
发现一个与 std :: begin
。
b
namespace foo {
struct A {int begin; };
struct B {using end = int; };
class C {int * begin(); int * end(); }; //无法访问
struct D {int * begin(int); int * end();};
struct E {};
模板< class T> int * begin(T&){return nullptr; }
模板< class T> int * end(T&){return nullptr; }
}
foo :: A a; foo :: B b; foo :: C c; foo :: D d; foo :: E e;
我想 magic_begin(a)
/ magic_begin(b)中
/ magic_begin(c)中
/ magic_begin(d)
是一个编译错误,并且 magic_begin(e)
返回(int *)nullptr
。
下面的SFINAE友好的方法似乎按照需要工作(见下面例外):
#include< type_traits>
namespace detail {
struct empty {};
模板< typename T>
using base = std :: conditional_t< std :: is_class< T> {}&&不是std :: is_final< T> {},
T,空>;
struct P1 {typedef int begin,end;};
模板< typename U> $ b $ struct struct TestMemType:base< U> ;, P1 {
template< typename T = TestMemType,typename = typename T :: begin>
static std :: true_type test_begin(int);
模板< typename T = TestMemType,typename = typename T :: end>
static std :: true_type test_end(int);
static std :: false_type test_begin(float),test_end(float);
};
模板< typename T>
constexpr bool hasMember =!decltype(TestMemType< T> :: test_begin(0)){}
|| !decltype(TestMemType< T> :: test_end(0)){};
//!步骤1
模板< typename T,std :: size_t N>
constexpr auto begin(int,T(& a)[N]){return a;}
模板< typename T,std :: size_t N>
constexpr auto end(int,T(& a)[N]){return a + N;}
//!第二步 - 这个过载比上面的要少。
模板< typename T>
constexpr auto begin(int,T& a) - > decltype(a.begin()){return a.begin();}
template< typename T>
constexpr auto end(int,T& a) - > decltype(a.end()){return a.end();}
//!步骤3
命名空间nested_detail {
void begin(),end();
模板< typename T>
constexpr auto begin_(T& a) - > decltype(begin(a)){return begin(a);}
模板< typename T>
constexpr auto end_(T& a) - > decltype(end(a)){return end(a);}
}
模板< typename T,typename = std :: enable_if_t< not hasMember< std :: decay_t< T>>> ;>
constexpr auto begin(float,T& a) - > decltype(nested_detail :: begin_(a))
{return nested_detail :: begin_(a);}
模板< typename T,typename = std :: enable_if_t< not hasMember< std :: decay_t< T>>>>
constexpr auto end(float,T& a) - > decltype(nested_detail :: end_(a))
{return nested_detail :: end_(a);}
}
模板< typename T>
constexpr auto magic_begin(T& a) - > decltype(detail :: begin(0,a))
{return detail :: begin(0,a);}
模板< typename T>
constexpr auto magic_end(T& a) - > decltype(detail :: end(0,a))
{return detail :: end(0,a);}
演示 。请注意,GCCs查找被打破,因为它不考虑在 TestMemType :: test_end / begin
中的 typename T :: begin
code>。可以在 此处 找到解决方法草图。
步骤2中的检查要求类类型是可派生的,这意味着此方法不能正确处理 final
类或联合 - 如果这些成员具有名称 begin
/ end
的无法访问的成员。
Consider the specification of the range-based for loop's begin-expr and end-expr (N4140 [stmt.ranged]/p1). Given a range __range
of type _RangeT
,
begin-expr and end-expr are determined as follows:
- if
_RangeT
is an array type, begin-expr and end-expr are__range
and__range + __bound
, respectively, where__bound
is the array bound. If_RangeT
is an array of unknown size or an array of incomplete type, the program is ill-formed;- if
_RangeT
is a class type, the unqualified-idsbegin
andend
are looked up in the scope of class_RangeT
as if by class member access lookup (3.4.5), and if either (or both) finds at least one declaration, begin-expr and end-expr are__range.begin()
and__range.end()
, respectively;- otherwise, begin-expr and end-expr are
begin(__range)
andend(__range)
, respectively, wherebegin
andend
are looked up in the associated namespaces (3.4.2). [ Note: Ordinary unqualified lookup (3.4.1) is not performed. —end note ]
Is it possible to simulate this exact behavior in ordinary C++ code? i.e., can we write a magic_begin
and a magic_end
function template such that
for(auto&& p : range_init) { /* statements */ }
and
{
auto&& my_range = range_init;
for(auto b = magic_begin(my_range), e = magic_end(my_range); b != e; ++b){
auto&& p = *b;
/* statements */
}
}
always have the exact same behavior?
Non-answers include qualified calls to std::begin
/std::end
(doesn't handle the third bullet, among other things) and using std::begin; begin(range);
because, among other things, that is ambiguous if ADL for begin
finds an overload that's equally good as std::begin
.
For illustration, given
namespace foo {
struct A { int begin; };
struct B { using end = int; };
class C { int* begin(); int *end(); }; // inaccessible
struct D { int* begin(int); int* end();};
struct E {};
template<class T> int* begin(T&) { return nullptr; }
template<class T> int* end(T&) { return nullptr; }
}
foo::A a; foo::B b; foo::C c; foo::D d; foo::E e;
I want magic_begin(a)
/magic_begin(b)
/magic_begin(c)
/magic_begin(d)
to be a compile error, and magic_begin(e)
to return (int*)nullptr
.
The following SFINAE-friendly approach seems to work as desired (see below for exceptions):
#include <type_traits>
namespace detail {
struct empty {};
template <typename T>
using base = std::conditional_t<std::is_class<T>{} && not std::is_final<T>{},
T, empty>;
struct P1 {typedef int begin, end;};
template <typename U>
struct TestMemType : base<U>, P1 {
template <typename T=TestMemType, typename=typename T::begin>
static std::true_type test_begin(int);
template <typename T=TestMemType, typename=typename T::end>
static std::true_type test_end(int);
static std::false_type test_begin(float), test_end(float);
};
template <typename T>
constexpr bool hasMember = !decltype(TestMemType<T>::test_begin(0)){}
|| !decltype(TestMemType<T>::test_end(0)){};
//! Step 1
template <typename T, std::size_t N>
constexpr auto begin(int, T(&a)[N]) {return a;}
template <typename T, std::size_t N>
constexpr auto end(int, T(&a)[N]) {return a+N;}
//! Step 2 - this overload is less specialized than the above.
template <typename T>
constexpr auto begin(int, T& a) -> decltype(a.begin()) {return a.begin();}
template <typename T>
constexpr auto end(int, T& a) -> decltype(a.end()) {return a.end();}
//! Step 3
namespace nested_detail {
void begin(), end();
template <typename T>
constexpr auto begin_(T& a) -> decltype(begin(a)) {return begin(a);}
template <typename T>
constexpr auto end_(T& a) -> decltype(end(a)) {return end(a);}
}
template <typename T, typename=std::enable_if_t<not hasMember<std::decay_t<T>>>>
constexpr auto begin(float, T& a) -> decltype(nested_detail::begin_(a))
{return nested_detail::begin_(a);}
template <typename T, typename=std::enable_if_t<not hasMember<std::decay_t<T>>>>
constexpr auto end(float, T& a) -> decltype(nested_detail::end_(a))
{return nested_detail::end_(a);}
}
template <typename T>
constexpr auto magic_begin(T& a) -> decltype(detail::begin(0, a))
{return detail::begin(0, a);}
template <typename T>
constexpr auto magic_end (T& a) -> decltype(detail::end (0, a))
{return detail:: end(0, a);}
Demo. Note that GCCs lookup is broken as it doesn't consider non-type names for typename T::begin
in TestMemType::test_end/begin
. A workaround sketch can be found here.
The check in step 2 requires that the class type be derivable, which implies that this method doesn't properly work with final
classes or unions - if those have an inaccessible member with name begin
/end
.
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