In the example below, why line 20 causes the error described from line 27 to 30?

Calling exec1 in line 33 works fine.

#include <cstdint>
#include <functional>
#include <iostream>
#include <tuple>
#include <type_traits>

template <typename... t_fields>
void exec0(std::function<std::tuple<t_fields...>()> generate,
           std::function<void(t_fields &&...)> handle) {
  std::tuple<t_fields...> _tuple{generate()};
  std::apply(handle, std::move(_tuple));
}

template <typename t_generate, typename t_function>
void exec1(t_generate generate, t_function handle) {
  auto _tuple{generate()};
  std::apply(handle, std::move(_tuple));
}

int main() {
  auto _h = [](uint32_t u) -> void { std::cout << "u = " << u << '\n'; };

  auto _g = []() -> std::tuple<uint32_t> { return std::tuple<uint32_t>{456}; };

  //  exec0<uint32_t>(_g, _h);
  /*
main.cpp:25:3: error: no matching function for call to 'exec0'
main.cpp:8:6: note: candidate template ignored: could not match
'function<tuple<unsigned int, type-parameter-0-0...> ()>' against '(lambda at
/var/tmp/untitled002/main.cpp:23:13)'
  */

  exec1(_g, _h);

  return 0;
}

g --version replies:

g   (Ubuntu 9.4.0-1ubuntu1~20.04.1) 9.4.0
Copyright (C) 2019 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

CodePudding user response：

Even though you specified <uint32_t> as a template argument, the compiler seems to try to deduce more elements for the parameter pack, fails to do so (because the type of a lambda is not std::function<...>), and becomes upset.

You need to somehow inhibit template argument deduction.

Either call it as exec0<uint32_t>({_g}, {_h});, or wrap parameter types in std::type_identity_t<...> (or, pre-C 20, std::enable_if_t<true, ...>).

Then the compiler will accept your uint32_t as the only type in the pack, and won't try to add more types.