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Retrieving refusal from repeating a template iterator

I need to get a link to a link iterator. However, my compiler is choking on this code:

template <typename InputIterator> size_t iLongestBegin(InputIterator first, InputIterator last)
{
    typedef typename std::iterator_traits<InputIterator>::reference SequenceT;
        //Problem is next line
    typedef typename std::iterator_traits<typename SequenceT::iterator>::reference T;
    for(size_t idx; idx < first->length(); idx++)
    {
        T curChar = (*first)[idx];
        for (InputIterator cur = first; cur != last; cur++)
        {
            if (cur->length() < idx)
                return idx;
            if (_tolower(cur->at(idx)) != _tolower(curChar))
                return idx;
        }
    }
    return first->length();
}

Any ideas on how to fix this? Error

error C2825: 'SequenceT': must be a class or namespace when followed by '::'

Thanks! Billy3

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4 answers


Actually, I just decided :)

The problem is that SequenceT is a reference and not a type. Since you cannot normally use a reference type address, the compiler will not generate iterators for it. I need to use value_type instead of reference:



template <typename InputIterator> size_t iLongestBegin(InputIterator first, InputIterator last)
{
    typedef typename std::iterator_traits<InputIterator>::reference SequenceT;
    typedef typename std::iterator_traits<std::iterator_traits<InputIterator>::value_type::iterator>::reference T;
    for(size_t idx; idx < first->length(); idx++)
    {
        typename T curChar = (*first)[idx];
        for (InputIterator cur = first; cur != last; cur++)
        {
            if (cur->length() < idx)
                return idx;
            if (_tolower(cur->at(idx)) != _tolower(curChar))
                return idx;
        }
    }
    return first->length();
}
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You were unable to specify SequenceT

in the template argument list - Where is it defined? Or maybe you should indicate that it is type c typename SequenceT

.



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You need to write typename SequenceT::iterator

instead SequenceT::iterator

. This is because it SequenceT

is a type derived from your template parameters ("dependent type" in standard linguistic), and iterator

is a nested type in SequenceT

, not a function or variable. When both of these things are true, the compiler cannot figure out what you mean and it must be said that SequenceT::iterator

it is type c typename

.

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The following compilations with g ++ 4.4.0:

#include <iterator>
using namespace std;

template <typename InputIterator> size_t iLongestBegin(InputIterator first, InputIterator last) {
   typedef typename std::iterator_traits<InputIterator>::reference SequenceT;
   typedef typename std::iterator_traits<typename SequenceT::iterator>::reference T;
   for(size_t idx; idx < first->length(); idx++)
   {
      T curChar = (*first)[idx];
      for (InputIterator cur = first; cur != last; cur++)
      {
         if (cur->length() < idx)
             return idx;
         if (_tolower(cur->at(idx)) != _tolower(curChar))
             return idx;
      }
   }
    return first->length();
}
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