boost::mpl::vector - 获取类型的基本偏移量 [英] boost::mpl::vector - getting to a type's base-offset

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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