Assign a pointer to a lambda function countaining variables passed by reference - c++

Question

I would like to assign a pointer to a lambda function, in which the lambda function is taking variables passed by reference, not by value.
int main() {
// what I can do, but not quite what I want
auto funcy = [](const double i ) {
std:: cout << "this is i: " << i << std::endl;
};
void(*lptr)(double); // OK
lptr = funcy;
lptr(1);
// TODO: make a variable that points to this lambda work
auto funcy2 = [](const double &i ) {
std:: cout << "this is i: " << i << std::endl;
};
void(*lptr2)(double *); // BROKE
lptr2 = funcy2;
lptr2(1);
return 0;
}
Is it possible to do this?
Thanks for your time.
Edit: This post is different from Passing capturing lambda as function pointer because I have no idea what that one is saying.

Answer 1

Answered by #sweenish in the comments. The "function signature" has to match. It's actually "pretty straightforward".
void(*lptr2)(const double&); // FIXED
lptr2 = funcy2;
lptr2(5);

Answer 2

The best would be that the type of the function pointer would match the type of the lambda function. This will be much more simple and is most likely the solution that you want.
By using pointers everywhere
auto funcy2 = [](double *i) {
std::cout << "this is i: " << *i << std::endl;
};
void(*lptr2)(double *); // matches
lptr2 = funcy2;
double i = 1;
lptr2(&i);
... or even better, references everywhere:
auto funcy2 = [](const double &i) {
std::cout << "this is i: " << i << std::endl;
};
void(*lptr2)(const double &); // works, the type match the lambda parameter
lptr2 = funcy2;
lptr2(1);
If you really want to assign a lambda to a mismatched type, you'll have to wrap it. Since there's a conversion to do, you can do it using another lambda:
static auto funcy2 = [](const double &i) {
std::cout << "this is i: " << i << std::endl;
};
auto funcy3 = [](double* i) {
funcy2(&i); // convert the pointer to a reference
};
void(*lptr2)(double *); // BROKE
lptr2 = funcy2;
lptr2(1);