Sar*_*ran 7 c++ icc decltype variadic-templates c++11
我正在为由任意数量的char标签参数化的表达式编写模板.
给定参数列表,工厂函数返回不同类型的表达式,具体取决于是否存在相同类型的两个参数或它们是否唯一.
一个具体的例子:假设这A是一个"可标记的"对象,其operator()重载生成一个?Expression<...>.我们a, b, ...将其声明为标签LabelName<'a'>, LabelName<'b'>, ....然后A(a,b,c,d)会产生一个UniqueExpression<'a','b','c','d'>,而反过来A(a,c,b,c)会产生一个RepeatedExpression<'a','c','b','c'>.
为了实现这一点,我不得不?Expression用auto和定义工厂函数decltype.此外,decltype必须级联到另一个,decltype直到元程序完成通过参数的递归并且最终决定返回类型.作为一个例子,我已经为工厂方法隔离了一个相当小的代码.
template <typename... T> struct TypeList { };
template <char C> struct LabelName { };
template <typename... T> class UniqueExpression
{
// Contains implementation details in actual code
};
template <typename... T> class RepeatedExpression
{
// Contains implementation details in actual code
};
class ExpressionFactory {
private:
template <char _C, typename... T, typename... _T>
static UniqueExpression<T...>
_do_build(TypeList<T...>,
TypeList<LabelName<_C>>,
TypeList<>,
TypeList<_T...>)
{
return UniqueExpression<T...> ();
}
template <char _C, typename... T, typename... _T1, typename... _T2, typename... _T3>
static RepeatedExpression<T...>
_do_build(TypeList<T...>,
TypeList<LabelName<_C>, _T1...>,
TypeList<LabelName<_C>, _T2...>,
TypeList<_T3...>)
{
return RepeatedExpression<T...> ();
}
template <char _C1, char _C2, typename... T, typename... _T1, typename... _T2, typename... _T3>
static auto
_do_build(TypeList<T...>,
TypeList<LabelName<_C1>, _T1...>,
TypeList<LabelName<_C2>, _T2...>,
TypeList<_T3...>)
-> decltype(_do_build(TypeList<T...>(),
TypeList<LabelName<_C1>, _T1...>(),
TypeList<_T2...>(),
TypeList<_T3..., LabelName<_C2>>()))
{
return _do_build(TypeList<T...>(),
TypeList<LabelName<_C1>, _T1...>(),
TypeList<_T2...>(),
TypeList<_T3..., LabelName<_C2>>());
}
template <char _C1, char _C2, typename... T, typename... _T1, typename... _T2>
static auto
_do_build(TypeList<T...>,
TypeList<LabelName<_C1>, LabelName<_C2>, _T1...>,
TypeList<>,
TypeList<LabelName<_C2>, _T2...>)
-> decltype(_do_build(TypeList<T...>(),
TypeList<LabelName<_C2>, _T1...>(),
TypeList<_T2...>(),
TypeList<>()))
{
return _do_build(TypeList<T...>(),
TypeList<LabelName<_C2>, _T1...>(),
TypeList<_T2...>(),
TypeList<>());
}
public:
template <char C, typename... T>
static auto
build_expression(LabelName<C>, T...)
-> decltype(_do_build(TypeList<LabelName<C>, T...>(),
TypeList<LabelName<C>, T...>(),
TypeList<T...>(),
TypeList<>()))
{
return _do_build(TypeList<LabelName<C>, T...>(),
TypeList<LabelName<C>, T...>(),
TypeList<T...>(),
TypeList<>());
}
};
可以像这样在程序operator()中调用工厂:(在实际程序中有另一个带有重载的类调用工厂)
int main()
{
LabelName<'a'> a;
LabelName<'b'> b;
...
LabelName<'j'> j;
auto expr = ExpressionFactory::build_expression(a,b,c,d,e,f,g,h,i,j);
// Perhaps do some cool stuff with expr
return 0;
}
上面的代码按预期工作,并由GCC和英特尔编译器正确编译.现在,我明白编译器会花费更多的时间来执行递归模板推导,因为我增加了我使用的标签数量.
在我的计算机上,如果build_expression用一个参数调用,那么GCC 4.7.1平均需要大约0.26秒来编译.对于五个参数,编译时间扩展到大约0.29秒,对于十个参数,编译时间扩展到0.62秒.这一切都非常合理.
这个故事与英特尔编译器截然不同.ICPC 13.0.1在0.35秒内编译单参数代码,编译时间对于最多四个参数保持相当恒定.在五个参数中,编译时间最多为12秒,在六个参数下,它会在9600秒以上(即超过2小时40分钟)上升.不用说,我没有等到足以找出编译七参数版本需要多长时间.
立刻想到两个问题:
特别知道英特尔编译器编译递归的速度慢decltype吗？
有没有办法重写这个代码,以一种可能对编译器更友好的方式实现相同的效果？
Here is a stab at it. Instead of doing pairwise comparisons of each of the elements, I sort the list of types, then use a brain-dead unique algorithm to see if there are any duplicates.
I implemented merge sort on types, because it was fun. Probably a naive bubble sort would work better on reasonable number of arguments. Note that a bit of work would allow us to do a merge sort on long lists, and specialize for bubble sorts (or even insertion sorts) on short lists. I'm not up for writing a template quicksort.
This gives me a compile time boolean that says if there are duplicates in the list. I can then use enable_if to pick which overload I'm going to use.
Note that your solution involved n^2 layers of template recursion, at each stage of which the return type requires evaluating the type of a 1 step simpler class, and then the type returned also requires the same! If the Intel compiler memoization fails, you are talking exponential amounts of work.
我用一些助手增强了你们的一些课程。我让你的LabelNames 继承自std::integral_constant，因此我可以轻松地在编译时访问它们的值。这使得排序代码变得更容易一些。我还公开了enum重复且唯一的返回值，以便我可以printf对结果进行简单的调试。
这项工作的大部分内容是编写合并排序——是否有我们可以使用的标准编译时类型排序？
#include <type_traits>
#include <iostream>
template <typename... T> struct TypeList { };
// NOTE THIS CHANGE:
template <char C> struct LabelName:std::integral_constant<char, C> {};
template <typename... T> class UniqueExpression
{
// Contains implementation details in actual code
public:
enum { is_unique = true };
};
template <typename... T> class RepeatedExpression
{
// Contains implementation details in actual code
public:
enum { is_unique = false };
};
// A compile time merge sort for types
// Split takes a TypeList<>, and sticks the even
// index types into Left and odd into Right
template<typename T>
struct Split;
template<>
struct Split<TypeList<>>
{
typedef TypeList<> Left;
typedef TypeList<> Right;
};
template<typename T>
struct Split<TypeList<T>>
{
typedef TypeList<T> Left;
typedef TypeList<> Right;
};
// Prepends First into the TypeList List.
template<typename First, typename List>
struct Prepend;
template<typename First, typename... ListContents>
struct Prepend<First,TypeList<ListContents...>>
{
typedef TypeList<First, ListContents...> type;
};
template<typename First, typename Second, typename... Tail>
struct Split<TypeList<First, Second, Tail...>>
{
typedef typename Prepend< First, typename Split<TypeList<Tail...>>::Left>::type Left;
typedef typename Prepend< Second, typename Split<TypeList<Tail...>>::Right>::type Right;
};
// Merges the sorted TypeList<>s Left and Right to the end of TypeList<> MergeList
template< typename Left, typename Right, typename MergedList=TypeList<> >
struct Merge;
template<typename MergedList>
struct Merge< TypeList<>, TypeList<>, MergedList >
{
typedef MergedList type;
};
template<typename L1, typename... Left, typename... Merged>
struct Merge< TypeList<L1, Left...>, TypeList<>, TypeList<Merged... >>
{
typedef TypeList<Merged..., L1, Left...> type;
};
template<typename R1, typename... Right, typename... Merged>
struct Merge< TypeList<>, TypeList<R1, Right...>, TypeList<Merged...> >
{
typedef TypeList<Merged..., R1, Right...> type;
};
template<typename L1, typename... Left, typename R1, typename... Right, typename... Merged>
struct Merge< TypeList<L1, Left...>, TypeList<R1, Right...>, TypeList<Merged...>>
{
template<bool LeftIsSmaller, typename LeftList, typename RightList, typename MergedList>
struct MergeHelper;
template<typename FirstLeft, typename... LeftTail, typename FirstRight, typename... RightTail, typename... MergedElements>
struct MergeHelper< true, TypeList<FirstLeft, LeftTail...>, TypeList<FirstRight, RightTail...>, TypeList<MergedElements...> >
{
typedef typename Merge< TypeList<LeftTail...>, TypeList< FirstRight, RightTail... >, TypeList< MergedElements..., FirstLeft > >::type type;
};
template<typename FirstLeft, typename... LeftTail, typename FirstRight, typename... RightTail, typename... MergedElements>
struct MergeHelper< false, TypeList<FirstLeft, LeftTail...>, TypeList<FirstRight, RightTail...>, TypeList<MergedElements...> >
{
typedef typename Merge< TypeList<FirstLeft, LeftTail...>, TypeList<RightTail... >, TypeList< MergedElements..., FirstRight > >::type type;
};
typedef typename MergeHelper< (L1::value < R1::value), TypeList<L1, Left...>, TypeList<R1, Right...>, TypeList<Merged...> >::type type;
};
// Takes a TypeList<T...> and sorts it via a merge sort:
template<typename List>
struct MergeSort;
template<>
struct MergeSort<TypeList<>>
{
typedef TypeList<> type;
};
template<typename T>
struct MergeSort<TypeList<T>>
{
typedef TypeList<T> type;
};
template<typename First, typename Second, typename... T>
struct MergeSort<TypeList<First, Second, T...>>
{
typedef Split<TypeList<First, Second, T...>> InitialSplit;
typedef typename MergeSort< typename InitialSplit::Left >::type Left;
typedef typename MergeSort< typename InitialSplit::Right >::type Right;
typedef typename Merge< Left, Right >::type type;
};
// Sorts a TypeList<T..>:
template<typename List>
struct Sort: MergeSort<List> {};
// Checks sorted TypeList<T...> SortedList for adjacent duplicate types
// return value is in value
template<typename SortedList>
struct Unique;
template<> struct Unique< TypeList<> >:std::true_type {};
template<typename T> struct Unique< TypeList<T> >:std::true_type {};
template<typename First, typename Second, typename... Tail>
struct Unique< TypeList< First, Second, Tail... > >
{
enum
{
value = (!std::is_same<First, Second>::value) &&
Unique< TypeList<Second, Tail...> >::value
};
};
// value is true iff there is a repeated type in Types...
template<typename... Types>
struct RepeatedType
{
typedef TypeList<Types...> MyListOfTypes;
typedef typename Sort< MyListOfTypes >::type MyListOfTypesSorted;
enum
{
value = !Unique< MyListOfTypesSorted >::value
};
};
// A struct that creates an rvalue trivial constructed type
// of any type requested.
struct ProduceRequestedType
{
template<typename Result>
operator Result() { return Result(); };
};
struct ExpressionFactory {
template<typename... T>
typename std::enable_if<
!RepeatedType<T...>::value,
UniqueExpression<T...>
>::type
build_expression(T...) const
{
return ProduceRequestedType();
};
template<typename... T>
typename std::enable_if<
RepeatedType<T...>::value,
RepeatedExpression<T...>
>::type
build_expression(T...) const
{
return ProduceRequestedType();
};
};
// Simple testing code for above:
int main()
{
auto foo1 = ExpressionFactory().build_expression( LabelName<'a'>(), LabelName<'b'>(), LabelName<'a'>() );
typedef decltype(foo1) foo1Type;
if (foo1Type::is_unique)
std::cout << "foo1 is unique\n";
else
std::cout << "foo1 is repeated\n";
auto foo2 = ExpressionFactory().build_expression( LabelName<'q'>(), LabelName<'a'>(), LabelName<'b'>(), LabelName<'d'>(), LabelName<'t'>(), LabelName<'z'>() );
typedef decltype(foo2) foo2Type;
if (foo2Type::is_unique)
std::cout << "foo2 is unique\n";
else
std::cout << "foo2 is repeated\n";
}
另外，我想对您的代码进行批评：模板编程就是编程——您的类型名称相当于在函数中使用“i1”到“i9”作为整数变量。当做一些不平凡的事情时，给你的类型命名有意义的名字。
它如何在 Intel 上编译？