迭代加速融合::矢量

问题描述

我想遍历使用一个boost ::融合表达载体：

I'm trying to iterate over a boost::fusion vector using:

typedef typename fusion::result_of::begin<T>::type t_iter;
  std::cout << distance(begin(t), end(t)) << std::endl;
  for(t_iter it = begin(t); it != end(t); next(it)){
    std::cout<<deref(it)<<std::endl;
  }

距离COUT声明给了我一个有限长度（2），但环似乎会无限期地运行。

The distance cout statement gives me a finite length (2), however the loop seems to run indefinitely.

任何意见多少AP preciated！

Any advice much appreciated!

推荐答案

您不能只是一个迭代的融合的载体这样，每个迭代器的类型可能会比previous不同一个（通常是这样）。我想这就是为什么你没有 IT =下一个（它）在code，它会给编译错误。

You can't just iterate a Fusion vector like that, the type for each iterator may be different than the previous one (and usually is). I guess that's why you don't have it = next(it) in your code, it would give a compilation error.

您可以使用的boost ::融合:: for_each的此，与打印每一个元素到标准输出函数对象在一起：

You could use boost::fusion::for_each for this, together with a function object that prints each element to the standard output:

struct print
{
    template< typename T >
    void operator()( T& v ) const
    {
        std::cout << v;
    }
};

...

boost::fusion::for_each( t, print() );

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