David Abrahams
Are you asking about runtime lambda expressions as in the Boost Lambda library, or compile-time lambda expressions as in MPL?
-- Dave Abrahams Boost Consulting www.boost-consulting.com
I was talking about MPL because that is the only one I have studied at the moment. What is the difference between the two outside of the obvious runtime/compile-time difference? I was trying to figure out how a particular placeholder gets its arguments to choose from when deciding the argument (argument 1 for _1 etc.). Thank you. Joseph L. McCay
Joe McCay
David Abrahams
writes: Are you asking about runtime lambda expressions as in the Boost Lambda library, or compile-time lambda expressions as in MPL?
-- Dave Abrahams Boost Consulting www.boost-consulting.com
I was talking about MPL because that is the only one I have studied at the moment. What is the difference between the two outside of the obvious runtime/compile-time difference?
Well, the syntax. MPL's compile-time lambdas never use parens as in the _1(5) you cited.
I was trying to figure out how a particular placeholder gets its arguments to choose from when deciding the argument (argument 1 for _1 etc.).
It's just in the definition of the placeholders
template <int> struct arg;
typedef arg<1> _1;
typedef arg<2> _2;
…
template <> struct arg<N>
{
template <
class A1 = void_
, class A2 = void_
…
, class AM = void_
>
struct apply
{
typedef AN type;
};
};
and the way arguments are passed through from apply_wrap.
apply_wrap<_2, int, int*, int**, int***>::type => int*
Maybe this simplified definition of lambda would help:
template <class T>
struct lambda { typedef T type; };
template < template <class> class F, class A1 >
struct lambda< F<A1> >
: if_<
is_placeholder_expression< F<A1> >
, bind< quote1<F>, typename lambda<A1>::type >
, F<A1>
>
{};
template < template
participants (2)
-
David Abrahams
-
Joe McCay