Parameter pack expansion with lambda

Question

I am trying to instantiate a function table (to emulate switch)

  template<size_t ... N>
  int f(std::index_sequence<N...>, int k)
  {
    static auto f_table = { []() { return N; }... };
    auto f = f_table.begin() + k;
    assert((*f)() == k);
    return (*f)();
  }

which fails with error: parameter packs not expanded with ‘...’:

I can get by with an extra wrapper function but why does lambda fail and is there a workaround?

GCC? Ancient? https://gcc.gnu.org/bugzilla/show_bug.cgi?id=64488 — tevemadar, Sep 06 '19 at 23:26
This [works](https://gcc.godbolt.org/z/CTXf0P) in new GCC with minor modifications. — HolyBlackCat, Sep 06 '19 at 23:27
https://stackoverflow.com/questions/40752568/expanding-parameter-pack-into-lambda-with-fold-expression-gcc-vs-clang had it in 7.0 — tevemadar, Sep 06 '19 at 23:33
@HolyBlackCat: …Because the lambdas have different types otherwise, if that’s not obvious to everyone. — Davis Herring, Sep 07 '19 at 01:12
you need at least G++ 8.1 to make this work. could look into using a different ppa, or build from source. Too bad JonathonF took his ppa's offline. — JHBonarius, Jan 22 '22 at 09:13

score 1 · Answer 1 · answered Jan 22 '22 at 09:00

It was a bug that GCC doesn't expand template parameter pack that appears in a lambda-expression, which was fixed in GCC 8 eventually.

The other issue with the code as also mentioned in the comments is that every lambda in the list { []() { return N; }... } has its own distinct type, so one has to convert them into function pointers with + operator first:

#include <utility>
#include <cassert>

template<std::size_t ... N>
int f(std::index_sequence<N...>, std::size_t k) {
   static auto f_table = { +[]() { return N; }... };
   auto f = f_table.begin() + k;
   assert((*f)() == k);
   return (*f)();
}

int main() {
    f(std::make_index_sequence<3>{}, 2);
}

Demo: https://gcc.godbolt.org/z/jWrfn7d6W

Parameter pack expansion with lambda

1 Answers1