提高:: MPL :: vector的 - 让一个类型的基偏移 [英] boost::mpl::vector - getting to a type's base-offset

问题描述

时有可能得到在一个偏移MPL ::矢量执行后MPL ::找到＆lt;序列，类型＆GT; 就可以了？

Is it possible to get at the offset of a mpl::vector after performing a mpl::find<seq,type> on it ?

换句话说我想要做的编译时间equavalent：

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;
}

如果做不到这一点，我的类型 MPL ::矢量有一个 type_trait＆LT; T＆GT; ::序常量硬 - codeD，我想避免这种可能的话

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，我也创建一个的boost ::变种的载体，我看我可以通过执行获得的序运行时函数变种::其中（）。然而，这需要我创建的默认初始化值的虚拟对象。这是相当uggly。如果你知道有变做了一些其他的方式，这将是一个解决我的问题也是如此。

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

如果你正在寻找的是一种的indexOf特点，我想从Boost.MPL文档有关的例子查找将这样的伎俩：

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 );

这篇关于提高:: MPL :: vector的 - 让一个类型的基偏移的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！