Christopher Schmidt wrote:
Fusion uses boost::result_of to determine the return type of a call
expression. Therefore you just need to define a result-metafunction.
Thanks. It works. However two things bother me.
First there's the warning :
1>c:\program
files\boost\boost_1_39\boost\fusion\view\transform_view\detail\deref_impl.hpp(42)
: warning C4172: returning address of local variable or temporary
1> c:\program
files\boost\boost_1_39\boost\fusion\view\transform_view\detail\deref_impl.hpp(41)
: while compiling class template member function 'const int
&boost::fusion::extension::deref_implboost::fusion::transform_view_iterator_tag::apply<Iterator>::call(const
Iterator &)'
1> with
1> [
1>
Iterator=boost::fusion::transform_view_iterator,0>,triple2>
1> ]
1> c:\program
files\boost\boost_1_39\boost\fusion\iterator\deref.hpp(53) : see
reference to class template instantiation
'boost::fusion::extension::deref_implboost::fusion::transform_view_iterator_tag::apply<Iterator>'
being compiled
1> with
1> [
1>
Iterator=boost::fusion::transform_view_iterator,0>,triple2>
1> ]
etc etc etc
Also, I see that transform returns a view. I need a proper vector that
is distinct from the input vector. Here's a bit of context. I am
building a recipient for one row in a database query. I have a vector
of Exprs, one for each column in the select. Expr has a type member. I
want to make a vector with an element for each Expr, initialized to the
type's default value (as per type()).
I.e. something like this :
struct f { /* ... */ template<typename Expr> operator ()(Expr) {
return typename Expr::type(); }
typedef result_of::transform::type row_type;
row_type row;
transform(select.exprs, row); // type of select.exprs is SelectList
std::deque results;
while (data) {
for_each(row, read_fun(...));
results.push_back(row);
}
Alas row_type is going to be a view :-/
J-L
J-L
