boost::mpl::vector - getting to a type#39;s base-offset(boost::mpl::vector - 获取类型的基本偏移量)
问题描述
在对 mpl::find<seq,type>
执行 mpl::find<seq,type>
之后,是否可以获得 mpl::vector
的偏移量?
Is it possible to get at the offset of a mpl::vector
after performing a mpl::find<seq,type>
on it ?
换一种说法,我想做以下等价的编译时间:
Put differently I want to do the compile time equavalent of:
#include <vector>
#include <algorithm>
#include <iostream>
int main()
{
typedef std::vector<int> v_type;
v_type v_int(3);
v_int[0] = 1;
v_int[1] = 2;
v_int[2] = 3;
v_type::iterator it= std::find( v_int.begin() ,v_int.end(),3);
std::cout << it - v_int.begin() << std::endl;
}
如果不这样做,我在 mpl::vector
中的类型有一个 type_trait<T>::ordinal
const 硬编码,如果可能的话,我想避免这种情况.
Failing this, my types in mpl::vector
have a type_trait<T>::ordinal
const hard-coded, I would like to avoid this if possible.
重要提示,我还从向量中创建了一个 boost::variant
,我发现我可以通过执行运行时函数 来获得序数变体::which()
.但是,这需要我创建一个具有默认初始化值的虚拟对象.这是相当难看的.如果您知道使用变体的其他方法,那也可以解决我的问题.
Important Note, I am also creating a boost::variant
from the vector, and I see I can get at the ordinal by performing a runtime function variant::which()
. However, this requires I create a dummy object with default-initialized values. This is quite uggly. If you know some other way of doing it with variant, that would be a solution to my problem as well.
推荐答案
如果您正在寻找的是一种 indexOf 功能,我猜 Boost.MPL 文档中关于 find
的示例将做这个伎俩:
If what you're looking for is a kind of indexOf feature, I guess the example from Boost.MPL doc concerning find
will do the trick:
typedef vector<char,int,unsigned,long,unsigned long> types;
typedef find<types,unsigned>::type iter;
BOOST_MPL_ASSERT(( is_same< deref<iter>::type, unsigned > ));
BOOST_MPL_ASSERT_RELATION( iter::pos::value, ==, 2 );
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本文标题为:boost::mpl::vector - 获取类型的基本偏移量
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