Obtaining function pointer to lambda?

Question

I want to be able to obtain a function pointer to a lambda in C++.

I can do:

int (*c)(int) = [](int i) { return i; };

And, of cours

Answer 1

The lambda code doesn't work for the same reason this doesn't work:

struct foo {
  operator int*() const {
    static int x;
    return &x;
  }
};

int* pint = foo{};
auto* pint2 = foo{}; // does not compile

or even:

template
void test(T*) {};
test(foo{});

The lambda has an operator that implicitly converts it to a (particular) function pointer, just like foo.

auto does not do conversion. Ever. Auto behaves like a class T parameter to a template function where its type is deduced.

As the type on the right hand side is not a pointer, it cannot be used to initialize an auto* variable.

Lambdas are not function pointers. Lambdas are not std::functions. They are auto-written function objects (objects with an operator()).

Examine this:

void (*ptr)(int) = [](auto x){std::cout << x;};
ptr(7);

it compiles and works in gcc (not certain if it is an extension, now that I think about it). However, what would auto* ptr = [](auto x){std::cout << x;} supposed to do?

However, unary + is an operator that works on pointers (and does nearly nothing to them), but not in foo or a lambda.

So

auto* pauto=+foo{};

And

auto* pfun=+[](int x){};

Both work, magically.