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yap/example/lazy_vector.cpp

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2.3 KiB
C++

#define BOOST_PROTO17_CONVERSION_OPERATOR_TEMPLATE
#include "expression.hpp"
#include <algorithm>
#include <cassert>
#include <iostream>
#include <vector>
// TODO: Turn this into a test that counts the number of allocations.
template <boost::proto17::expr_kind Kind, typename Tuple>
struct lazy_vector_expr;
struct take_nth
{
boost::proto17::terminal<double, lazy_vector_expr>
operator() (boost::proto17::terminal<std::vector<double>, lazy_vector_expr> const & expr);
std::size_t n;
};
template <boost::proto17::expr_kind Kind, typename Tuple>
struct lazy_vector_expr
{
using this_type = lazy_vector_expr<Kind, Tuple>;
static const boost::proto17::expr_kind kind = Kind;
Tuple elements;
BOOST_PROTO17_USER_BINARY_OPERATOR_MEMBER(plus, this_type, ::lazy_vector_expr)
BOOST_PROTO17_USER_BINARY_OPERATOR_MEMBER(minus, this_type, ::lazy_vector_expr)
auto operator[] (std::size_t n) const
{ return boost::proto17::evaluate(boost::proto17::transform(*this, take_nth{n})); }
};
boost::proto17::terminal<double, lazy_vector_expr>
take_nth::operator() (boost::proto17::terminal<std::vector<double>, lazy_vector_expr> const & expr)
{
double x = boost::proto17::value(expr)[n];
return boost::proto17::make_terminal<lazy_vector_expr, double>(std::move(x));
}
struct lazy_vector :
lazy_vector_expr<
boost::proto17::expr_kind::terminal,
boost::hana::tuple<std::vector<double>>
>
{
template <boost::proto17::expr_kind Kind, typename Tuple>
lazy_vector & operator+= (lazy_vector_expr<Kind, Tuple> const & rhs)
{
std::vector<double> & this_vec = boost::proto17::value(*this);
for (int i = 0, size = (int)this_vec.size(); i < size; ++i) {
this_vec[i] += rhs[i];
}
return *this;
}
};
int main ()
{
lazy_vector v1{{std::vector<double>(4, 1.0)}};
lazy_vector v2{{std::vector<double>(4, 2.0)}};
lazy_vector v3{{std::vector<double>(4, 3.0)}};
double d1 = (v2 + v3)[2];
std::cout << d1 << "\n";
v1 += v2 - v3;
std::cout << '{' << v1[0] << ',' << v1[1]
<< ',' << v1[2] << ',' << v1[3] << '}' << "\n";
// This expression is disallowed because it does not conform
// to the implicit grammar.
// (v2 + v3) += v1;
return 0;
}