Reputation: 32182
If I have a range and I want to transform adjacent pairs is there a boost range adaptor to do this?
for example
std::vector<int> a;
a.push_back(1);
a.push_back(2);
a.push_back(3);
auto b = a
| boost::range::transformed([](int x, int y){return x+y;});
and the output would be
3, 5
EDIT
I have made an attempt at a range adaptor
// Base class for holders
template< class T >
struct holder
{
T val;
holder( T t ) : val(t)
{ }
};
// Transform Adjacent
template <typename BinaryFunction> struct transformed_adjacent_holder
: holder
{
transformed_adjacent_holder(BinaryFunction fn) : holder<BinaryFunction>(fn)
};
template <typename BinaryFunction> transform_adjacent
(BinaryFunction fn)
{ return transformed_adjacent_holder<BinaryFunction>(fn); }
template< class InputRng, typename BinFunc>
inline auto Foo(const InputRng& r, const transformed_adjacent_holder<BinFunc> & f)
-> boost::any_range
< std::result_of(BinFunc)
, boost::forward_traversal_tag
, int
, std::ptrdiff_t
>
{
typedef boost::range_value<InputRng>::type T;
T previous;
auto unary = [&](T const & t) {auto tmp = f.val(previous, t); previous = t; return tmp; };
return r | transformed(unary);
}
However I don't know how to deduce the return type of the | operator. If I can do this then the adaptor is almost solved.
Upvotes: 1
Views: 218
Reputation: 393064
No. But you can use the range algorithm:
#include <boost/range/algorithm.hpp>
#include <boost/range/adaptors.hpp>
#include <vector>
#include <iostream>
using namespace boost::adaptors;
int main() {
std::vector<int> a { 1,2,3 };
auto n = a.size();
if (n > 0) {
auto out = std::ostream_iterator<int>(std::cout, "\n");
boost::transform(a, a | sliced(1, n), out, [](int x, int y) { return x + y; });
}
}
This uses the binary transform on a
and a slice of itself.
Prints
3
5
Upvotes: 2
Reputation: 7542
You can maintain a dummy variable
int prev=0;
auto b = a | transformed([&prev](int x){int what = x+prev;prev=x;return what;});
Upvotes: 2