问题
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;
}
The distance cout statement gives me a finite length (2), however the loop seems to run indefinitely.
Any advice much appreciated!
回答1:
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.
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() );
回答2:
fusion
is a wonderful library, and you should now that it is really different from what you use in every day C++ programs in multiple ways, it merge the power of compile time meta programming with runtime, for that you should now that there is no type that can handle all items in a fusion
container. What this means? it means that result_of::begin<T>::type
is not always a match of next(it)
so you can't use fusion
iterators in a for
like that.
The obvious problem in your code is that you ignore return value of next
and it will cause your code to run forever but you can't use it in it = next(it)
, since their type may vary!!
So what you should do?? You should use boost::fusion::for_each
for that purpose
回答3:
next
doesn't actually advance the iterator, it just returns the next one.
This can be seen in the docs, as the function next
takes a constant argument, meaning it can't possibly actually modify the iterator:
template<
typename I
>
typename result_of::next<I>::type next(I const& i);
^^^^^
回答4:
The problem is that inside the loop you are dereferencing your iterator. When you apply next
on it, it means nothing and that's why your loop runs forever.
来源:https://stackoverflow.com/questions/13094535/iterating-over-boost-fusionvector