...one of the most highly
regarded and expertly designed C++ library projects in the
world.
— Herb Sutter and Andrei
Alexandrescu, C++
Coding Standards
boost::proto::lazy::impl
// 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;
};
impl
public
types
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 functionsresult_type operator()(typename impl::expr_param expr, typename impl::state_param state, typename impl::data_param data) const;
behaves as follows:
proto::lazy
<T>::impl<Expr,State,Data>::operator()
If T
if of the form
O(A0,…An)
, then let O'
be boost::result_of<
and let proto::make
<O>(Expr, State, Data)>::typeT'
be O'(A0,…An)
.
If T
if of the form
O(A0,…An ...)
, then let O'
be boost::result_of<
and let proto::make
<O>(Expr, State, Data)>::typeT'
be O'(A0,…An ...)
.
Otherwise, let T'
be boost::result_of<
.
proto::make
<T>(Expr, State, Data)>::type
Returns: |
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