How to prevent compilation of passed lambda, if arguments are not references - c++

Question

In one of my projects I'm using a small utility function, which takes a Message struct and a lambda function, that modifies this message struct.
Now, I unintentionally passed a lambda without the necessary reference &. It perfectly compiles, but doesn't gave the desired output.
As for me, there should be one of the two following behaviors:
Forgetting to write auto&, but just auto should lead to compilation errors
Writing just auto should be interpreted as auto&.
It is possible to prevent compilation in case of a missing & or even better to interpret auto as auto& automatically?
#include <iostream>
#include <functional>
#include <boost/variant.hpp>
struct Message {
int x;
int y;
};
void changeMessage(Message& m, const std::function<void(Message&)>& messageModifier) {
std::cout << "Message before:" << m.x << " " << m.y << "\n";
messageModifier(m);
std::cout << "Message after:" << m.x << " " << m.y << "\n";
}
int main(int, char**) {
{
std::function<void(int&)> f = [](int&) {};
std::function<void(int)> g = [](int) {};
f = g; // This compiles.
}
{
std::function<void(int&)> f = [](int&) {};
std::function<void(int)> g = [](int) {};
//g = f; // This does not compile. Makes perfect sense.
}
Message m{ 10,20 };
{
changeMessage(m, [](auto m) { m.x++; m.y--; }); // User unintentionally forgot &! Can I prevent this from compilation?
std::cout << "Message outside: " << m.x << " " << m.y << "\n";
}
{
changeMessage(m, [](auto& m) { m.x++; m.y--; });
std::cout << "Message outside: " << m.x << " " << m.y << "\n";
}
}

Answer 1

One way to prevent passing Message by value (and auto itself is never a reference) is to disable copy construction:
struct Message {
Message() = default;
Message(const Message&) = delete;
int x;
int y;
};
Another solution suggested by #L. F. is to check that lambda doesn't accept rvalues:
template<class Fn>
void change_message(Message& m, Fn fn) {
static_assert(!std::is_invocable_v<Fn, Message&&>);
fn(m);
}