C++ variadic template empty argument specialization - c++

Question

What's the correct way to write a specialization for an empty argument variadic template. Take bellow code as an example:
#include <iostream>
#include <memory>
#include <tuple>
#include <functional>
#include <cassert>
using namespace std;
struct message {
int type;
};
struct X: message {
int payload;
X(): message{1} {
}
};
struct Y: message {
int payload;
Y(): message{2} {
}
};
struct Z: message {
int payload;
Z(): message{3} {
}
};
template<typename T>
constexpr int message_type = -1;
template<>
constexpr int message_type<X> = 1;
template<>
constexpr int message_type<Y> = 2;
template<>
constexpr int message_type<Z> = 3;
struct M {
int payload;
M(int payload): payload{ payload } {
}
};
template<typename P, typename T1, typename... Ts>
tuple<int, unique_ptr<M>> helper(unique_ptr<message> &msg, function<int(unique_ptr<T1>&)> fn1, function<int(unique_ptr<Ts>&)>... fn) {
if (msg->type == message_type<T1>) {
unique_ptr<T1> m(static_cast<T1*>(msg.release()));
auto result = fn1(m);
return {result, make_unique<M>(m->payload)};
} else {
return helper<void, Ts...>(msg, fn...);
}
}
template<typename P>
tuple<int, unique_ptr<M>> helper(unique_ptr<message> &msg) {
assert(false);
return {0, unique_ptr<M>()};
}
template<typename... Ts>
tuple<int, unique_ptr<M>> dispatch_msg(unique_ptr<message> &msg, function<int(unique_ptr<Ts>&)> ...fn) {
return helper<void, Ts...>(msg, fn...);
}
int main() {
auto *real_message = new Z;
real_message->payload = 101;
unique_ptr<message> msg(real_message);
auto [result, m] = dispatch_msg<X, Y, Z>(msg, [](auto &x) {
return x->payload + 1;
}, [](auto &y) {
return y->payload + 2;
}, [](auto &z) {
return z->payload + 3;
});
cout << result << '\n' << m->payload << endl;
return 0;
}
The helper function takes variadic template arguments. If it checked all given type arguments and failed. e.g. run to the empty arguments. I want to assert and stop the process.
The current code works but I'm wondering is there any straightforward way to write a specialization.
I simplified the core requirements into the code below:
template<typename T, typename... Ts>
void func(int val, T arg, Ts... args) {
if (condition_hold<T>(val)) {
return;
} else {
return func<Ts...>(val, args...);
}
}
template<>
void func(int val) {
assert(false);
}
int main() {
func<int, double, float>(100);
return 0;
}
Basically the func is checking against every given type whether a condition hold for the input val. If all check failed I want to do something, like the assert here. So I wrote a specialization takes empty argument, but this can't compile.

Answer 1

In C++17, you don't need to split parameter packs into head and tail in most cases. Thanks to fold expressions, many operations on packs become much easier.
// Some generic predicate.
template <typename T>
bool condition_hold(T) {
return true;
}
// Make this whatever you want.
void do_something_with(int);
template<typename... Ts>
auto func(int val, Ts... args) {
// Fold expression checks whether the condition is true for all
// elements of the parameter pack.
// Will be true if the parameter pack is empty.
if ((condition_hold(args) && ...))
do_something_with(val);
}
int main() {
// Ts type parameters are deduced to <float, float>.
func(100, 1.f, 2.f);
return 0;
}
To check whether the pack was empty and handle this case specially, you can do:
template<typename... Ts>
auto func(int val, Ts... args) {
if constexpr (sizeof...(Ts) == 0) {
// handle empty pack
}
else {
// handle non-empty pack
}
}
Your specialization couldn't have worked because func<> needs to take at least one parameter. A specialization such as
template<typename T>
void func<T>(int val);
Wouldn't be valid either, because it wold be a partial specialization which is only allowed for classes.
However, if the base template only takes a pack, we can fully specialize it:
template<typename... Ts>
void func(int val, Ts... args);
template<>
void func<>(int val);