Note
	Note that you can find everything that has been included and defined so far here.

This solution can still be improved. The foldl summarising the + <number> elements starts from 0 and once this is done, we add the value of the first <number> of the input to it in the first iteration. It would be more straightforward if foldl could use the value of the first <number> as the initial value of the "sum we have so far". Metaparse provides foldl_start_with_parser for this:

> #include <boost/metaparse/foldl_start_with_parser.hpp>

foldl_start_with_parser is almost the same as foldl. The difference is that instead of taking a starting value for the sum it takes a parser. First it parses the input with this parser and uses the value it returns as the starting value. Here is how we can implement our parser using it:

> using exp_parser11 = \
...> build_parser< \
...>   foldl_start_with_parser< \
...>     sequence<plus_token, int_token>, /* apply this parser repeatedly */ \
...>     int_token, /* use this parser to get the initial value */ \
...>     boost::mpl::quote2<sum_items> /* use this function to add a new value to the summary */ \
...>   > \
...> >;

copy-paste friendly version

This version of exp_parser uses foldl_start_with_parser. This implementation is more compact than the earlier versions. There is no sequence element in this: the first <number> is parsed by int_token and its value is used as the initial value for the summary. Let's try it out:

> exp_parser11::apply<BOOST_METAPARSE_STRING("1 + 2 + 3 + 4")>::type
mpl_::integral_c<int, 10>

It returns the same result as the earlier version but works differently. Here is a diagram showing how this implementation works: