Reputation: 891
#include <utility>
#include <tuple>
template < typename T, typename U >
void h(T, U)
{
}
template < typename... T, typename... U >
void f(std::tuple < T... > t, std::tuple < U... > u)
{
auto g = [&] < std::size_t... I > (std::index_sequence < I... >)
{
bool const r[]{((void)h(std::get < I >(t), std::get < I >(u)), false)...};
(void)r;
};
g(std::index_sequence_for < T... >());
}
int main()
{
f(std::make_tuple(0L, 0LL), std::make_tuple(0UL, 0ULL));
}
The above compiles with g++ test_templated_lambda.cpp -o test_templated_lambda -std=c++14
, but doesn't compile with clang++ test_templated_lambda.cpp -o test_templated_lambda -std=c++14
I know it is a GCC extension (Using template parameter in a generic lambda), but is there some way to do this without writing out g
as a free function
Upvotes: 4
Views: 935
Reputation: 157324
This is not possible without some external help; generic lambdas are the only form of template permitted within a function scope, and they cannot be specialized or overloaded (without some external helper, such as P0051R1 overload
).
It is possible to write a compile-time loop by embedding a recursive generic lambda in your function, but (a) you'll have to convert it to fix-point combinator form, and (b) terminating it is seriously ugly:
[&](auto&& g) {
g(g, std::integral_constant<std::size_t, 0u>{});
}([&](auto&& g, auto I) {
h(std::get<I>(t), std::get<I>(u));
std::get<I + 1 == sizeof...(T)>(std::make_pair(
[&](auto&& g) { g(g, std::integral_constant<std::size_t, I + 1>{}); },
[](auto&&){}))(g);
});
You're already using an external helper facility (std::index_sequence_for
), so why not write another? For example:
template<class F, std::size_t... I> auto apply_indexes(F&& f, std::index_sequence<I...>) {
return std::forward<F>(f)(std::integral_constant<std::size_t, I>{}...);
}
Usage:
auto g = [&](auto... I)
{
bool const r[]{((void)h(std::get<I>(t), std::get<I>(u)), false)...};
(void)r;
};
apply_indexes(g, std::index_sequence_for<T...>());
Upvotes: 3