Re: [Boost-users] Boost.MPL: Is it possible to apply a variable substitution without evaluating the lambda-expression?

AMDG Luke Simon wrote:

I am using BOOST_MPL_ASSERT to assert a lambda-expression applied to a sequence of types by folding the lambda-expression onto a sequence of types (see code and continue discussion below).

template struct mpl_assert { typedef typename apply::type type; BOOST_MPL_ASSERT((type)); };

template struct assert_for_all : foldr< mpl_assert, true_, xs

...

{};

<snip>

...but what I want to have printed is the type:

boost::is_same

Boost.MPL is already calculating such a type, because apply, bar> returns is_same::type, so the needed type is there... but does Boost.MPL provide a means of accessing it as opposed to accessing its ::type field?

MPL doesn't provide such a metafunction. Note that MPL doesn't actually calculate this in the process of evaluating lambda expressions, because of nesting. For example, is_same > does not ever use is_same You might try using protect instead of lambda in assert_for_all. In Christ, Steven Watanabe