Get error C2664 when parse xview as an argument to a function

[linggen]

@jmp75 @kou @minrk @eric-wieser @luk036
source code:

using XBufferType = xt::xbuffer_adaptor<double*, xt::no_ownership>;
using XAdaptType = xt::xarray_adaptor<XBufferType, xt::layout_type::row_major, std::vector<int32_t>>;
using XViewType = xt::xview<XAdaptType&,xt::xall<long unsigned int>, xt::xall<long unsigned int>, xt::xall<long unsigned int>>;

void factor(XViewType & view){
	std::cout << view(0, 0, 0) << "," << view(0, 1, 0) << std::endl;
}

std::shared_ptr<XAdaptType> stock_xarray_ = std::make_shared<XAdaptType>(xt::adapt(stock_factor_storage_.data(), stock_factor_storage_.size(), xt::no_ownership(), stock_factor_storage_.shape()

auto view = xt::view(*stock_xarray_, xt::all(), xt::all(), xt::all());
factor(view);

platform:
Windows VS 2019 ERROR
Return:
Error "C2664" void factor (XViewType &) ": unable to change parameter 1 from" view_Type" to "XViewType &"

[JohanMabille]

This means XViewType is not the typed returned by xt::view(*stock_array_, xt::all(), xt::all(), xt::all()).

If you want the argument type to be fully specified, you can "print" the type of the previous expression with the following:

template <class T>
struct DEBUG;

using view_type = decltype(view);
typename DEBUG<view_type>::type t;

This code will produce an error message containing the exact type of view.

If you can relax the constraint on the type (or if you need more generic code), but still want to ensure you pass a view on an XAdaptType&, whatever the slices are, you can turn factor into a template function:

template <class... SL>
void factor(xview<XAdaptType&, SL...>& view)
{
    ....
}