I have a template for a function that takes an output iterator argument. How can I use static_assert
to check that an instantiation uses an appropriate iterator? (Ie, both that it is an output iterator, and that it assigns elements of the correct type.)
#include <iostream>
#include <list>
#include <set>
template <class OutputIter>
void add_ints ( OutputIter iter )
{
static_assert ( something_goes_here,
"Arg must be an output iterator over ints" );
*iter++ = 1;
*iter++ = 2;
*iter++ = 3;
}
main()
{
// Insert iterator will add three elements.
std::set<int> my_set;
add_ints ( std::inserter ( my_set, my_set.end() ) );
for ( int i : my_set ) std::cout << i << "\n";
// Non-insert iterator will overwrite three elements.
std::list<int> my_list ( { 0, 0, 0 } );
add_ints ( my_list.begin() ) );
for ( int i : my_list ) std::cout << i << "\n";
#if 0
// Want nice compile error that container elements are not ints.
std::set<std::string> bad_set;
add_ints ( std::inserter ( bad_set, bad_set.end() ) );
#endif
#if 0
// Want nice compile error that argument is not an iterator.
class Foo {} foo;
add_ints ( foo );
#endif
}
OutputIterator
s are not required to have value types; their value_type
may well be void
, and in fact is void
for the purely output iterators from the standard library.
In your original question you checked for output_iterator_tag
, but you should not. There are plenty of perfectly mutable iterators that have a different category. std::vector<int>::iterator
's category is random_access_iterator_tag
, for instance.
Instead, check the well-formed-ness of the applicable expressions directly. All Iterator
s must support *r
and ++r
, and in addition OutputIterator
s must support *r = o
, r++
, and *r++ = o
, so:
template<class...>
struct make_void { using type = void; };
template<class... Ts>
using void_t = typename make_void<Ts...>::type;
template<class Iter, class U, class = void>
struct is_output_iterator_for : std::false_type {};
template<class Iter, class U>
struct is_output_iterator_for<
Iter, U,
void_t<decltype(++std::declval<Iter&>()),
decltype(*std::declval<Iter&>() = std::declval<U>()),
decltype(*std::declval<Iter&>()++ = std::declval<U>())>> : std::true_type {};