There is an sample for static_lambda use of static variant: /* begin */ int a = 5; int sub(int l, int r) { return l - r; } BOOST_STATIC_LAMBDA_DECLARE_EXPRESSION(int, expr_a, a); // equals to BOOST_STATIC_LAMBDA_DECLARE_EXPRESSION(BOOST_TYPEOF(&sub), expr_sub, &sub) BOOST_STATIC_LAMBDA_AUTO_EXPRESSION(expr_sub, &sub); int main() { BOOST_ASSERT(to_functor(expr_sub(constant<boost::mpl::int_<62> >(), expr_a)).static_invoke() == sub(62, a)); a = 10; BOOST_ASSERT(to_functor(expr_sub(constant<boost::mpl::int_<62> >(), expr_a))() == sub(62, a)); } /* end */ Any static variant be used must register first, because I can't get a variant value in compile time. Another way is using constant<>, like constant<boost::mpl::int_<62> >(). On 6/29/07, Alexander Nasonov <alnsn@yandex.ru> wrote:
Daniel Walker wrote:
Right. Actually, now that I think about it I'm getting a little confused about scenarios where this could save you space. I think they may be limited.
Scenarios for static stateless lambda expressions are limited, so it's not a surprise that scenarios where this could save you space are limited too ;)
We should carefully think about static _stateful_ lambda expressions. Some lambda expressions contain only constants and these expressions can be "complied" into static function. Though, it may lead to a confusion when a user pass a variable (not constant!) but static lambda uses a value passed first time.
int x = 0; (_1 + x).get_static_function<int>()(0); // returns 0 - ok x = 1; (_1 + x).get_static_function<int>()(0); // returns 0 - bad
Too bad. I have another use-case where stateful lambda could be useful but above problem doesn't give it any real chance :-(
-- Alexander Nasonov - http://nasonov.blogspot.com
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