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## Struct template impl

boost::proto::lazy::impl

## Synopsis

    // In header: <boost/proto/transform/lazy.hpp>

    template<typename Expr, typename State, typename Data>
    struct impl :  proto::transform_impl<Expr, State, Data> {
      // types
      typedef see-below result_type;

      // public member functions
      result_type operator()(typename impl::expr_param,
                             typename impl::state_param,
                             typename impl::data_param) const;
    };

## Description

### `impl` public types

1. typedef `see-below` result_type;

   `proto::lazy<T>::impl<Expr,State,Data>::result_type` is calculated as follows:

   - If `T` if of the form `O(A0,…An)`, then let `O'` be `boost::result_of<proto::make<O>(Expr, State, Data)>::type` and let `T'` be `O'(A0,…An)`.

   - If `T` if of the form `O(A0,…An ...)`, then let `O'` be `boost::result_of<proto::make<O>(Expr, State, Data)>::type` and let `T'` be `O'(A0,…An ...)`.

   - Otherwise, let `T'` be `boost::result_of<proto::make<T>(Expr, State, Data)>::type`.

   The result type is `boost::result_of<proto::call<T'>(Expr, State, Data)>::type` .

### `impl` public member functions

1.     result_type operator()(typename impl::expr_param expr,
                              typename impl::state_param state,
                              typename impl::data_param data) const;

   ``proto::lazy`<T>::impl<Expr,State,Data>::operator()` behaves as follows:

   - If `T` if of the form `O(A0,…An)`, then let `O'` be `boost::result_of<`proto::make`<O>(Expr, State, Data)>::type` and let `T'` be `O'(A0,…An)`.

   - If `T` if of the form `O(A0,…An ...)`, then let `O'` be `boost::result_of<`proto::make`<O>(Expr, State, Data)>::type` and let `T'` be `O'(A0,…An ...)`.

   - Otherwise, let `T'` be `boost::result_of<`proto::make`<T>(Expr, State, Data)>::type`.

   Returns: | `proto::call<T'>()(expr, state, data)`