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

This fails:

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

because the lambda is not a pointer. auto does not allow for conversions. Even though the lambda is convertible to something that is a pointer, that's not going to be done for you - you have to do it yourself. Whether with a cast:

auto *c = static_cast<int(*)(int)>([](int i){return i;});

Or with some sorcery:

auto *d = +[](int i) { return i; };

Answer 2

Especially when it can implicitly convert a unique, lambda type to a function pointer:

But it cannot convert it to "a function pointer". It can only convert it to a pointer to a specific function signature. This will fail:

int (*h)(float) = a;

Why does that fail? Because there is no valid implicit conversion from a to h here.

The conversion for lambdas is not compiler magic. The standard simply says that the lambda closure type, for non-capturing, non-generic lambdas, has an implicit conversion operator for function pointers matching the signature of its operator() overload. The rules for initializing int (*g)(int) from a permit using implicit conversions, and thus the compiler will invoke that operator.

auto doesn't permit using implicit conversion operators; it takes the type as-is (removing references, of course). auto* doesn't do implicit conversions either. So why would it invoke an implicit conversion for a lambda closure and not for a user-defined type?

Answer 3

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<class T>
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.