writable zip ranges are not possible?

问题

The following is failing:

#include <range/v3/view.hpp>
#include <range/v3/view/zip.hpp>
#include <range/v3/utility/iterator.hpp>

// ...

std::vector< std::tuple<int, std::string> > const data{
   {1,"a"},
   {2,"b"},
   {3,"c"}
};
std::vector<int> vi(data.size());
std::vector<std::string> vs(data.size());

using namespace ranges;
copy(data,  view::zip(vi,vs) ); // error

clang says

No matching function for call to object of type 'const 
ranges::v3::with_braced_init_args<ranges::v3::copy_fn>'

Assuming this is by design, why?

And, how can I do this obvious thing with ranges?

回答1:

1. copy takes an output iterator, not an output range. So you need to call begin on the zip view and turn it into an iterator.
2. With that fixed, you run into a separate problem. zipping two ranges produce a pair (well, a common_pair), but while tuples of two elements are assignable from pairs, pairs are not assignable from tuples of two elements. As a result, we can't do the equivalent of *zip_iterator = *data.begin(), and the concept check fails. If you make data a vector of pairs, then it would work.

来源：https://stackoverflow.com/questions/54858938/writable-zip-ranges-are-not-possible