When is *x also &x?

Via Ben Antonellis on CodeReview Stack Exchange:

template<class F>
void call_twice(F *f) {
    (*f)();
    (*f)();
}

static void one() {
    puts("one");
}
auto two = []() {
    puts("two");
};

int main() {
    call_twice(one);  // OK
    call_twice(two);  // ERROR - two is not a pointer
}

Okay, so let’s make two a pointer…

    call_twice(*two);  // OK

Wait… what?


Learners of C and C++ frequently have trouble with pointers — particularly the difference between & and * — particularly in C++, where both & and * can appear in declarations as well as in expressions. My tricks for overcoming this difficulty include:

  • Teach pointers first. In C++ we want to build the intuition that every object has an “address” in memory. The idea that we can represent memory addresses via the type system, just like we can represent natural numbers or Vehicles, is a hugely satisfying leap of logic. Teach references approximately five minutes later.

  • Teach the logic behind C-style declarations. int *p can mean “p is an int*,” and it can equally well mean “*p is an int.” int const *f() can mean “You aren’t allowed to modify the int,” or it can equally well mean “You aren’t allowed to modify *f().” (Of course this only works for pointers, not references; that’s another reason it’s important to reveal pointers early.)

  • “Follow the shooting star.” In an expression, both * and -> indicate that you’re following a pointer — you already have a pointer, and you’re following it to see what’s at the other end.

So in the code above, it’s unsurprising to see someone write two, have it fail to compile because could not match 'F *' against 'lambda', and then see them try to fix the bug by adding a star — *two — instead of an ampersand. What’s surprising is that adding the star works.


If you’ve seen my talk “Generic Lambdas From Scratch,” you know that a lambda is just syntactic sugar for a struct with a member operator(). When a lambda has no captures, then C++ gives the underlying struct one more method: a conversion function to the appropriate function pointer type. Our two is basically equivalent to the following struct:

using FuncPtr = void(*)();
struct TwoType {
    static void behavior() { puts("two"); }
    void operator()() const { behavior(); }
    operator FuncPtr() const { return &behavior; }
};
auto two = TwoType();

Usually we trigger this conversion to function pointer by doing something to the lambda that requires a scalar type. Commonly, idiomatically, the “coerce to scalar type” operator is unary +. (Unlike unary -, unary + works just fine on pointer types.)

auto pf = +[]() { puts("two"); };
static_assert(std::is_same_v<FuncPtr, decltype(pf)>);

You should also know that in expressions, functions very eagerly decay into function pointers, just as arrays eagerly decay into pointers-to-their-first-elements. For example, instead of return &behavior I could have written return behavior: the function behavior is happy to implicitly convert from void() to void(*)().

The unary * operator takes a function pointer and dereferences it to give me a function… which will then eagerly decay back to a function pointer if I do just about anything with it, including *-ing it again! So all of the following are equivalent:

void f();
auto pf1 = &f;
auto pf2 = f;
auto pf3 = *f;
auto pf4 = *******f;

In the initializer of pf4, f implicitly converts to a scalar function pointer so that we can * it; that gives us a reference to a function, which implicitly converts back to a function pointer so that we can * it the second time; and so on, arbitrarily many times.


That’s what happens in Ben’s code at the top of this post.

template<class F>
void call_twice(F *f);

static void one() { puts("one"); }
auto two = []() { puts("two"); };

int main() {
    call_twice(*one);  // OK
    call_twice(*two);  // OK
}

Unary * can’t apply directly to a function such as one, so one implicitly converts to a function pointer so that we can * it. That gives a function reference. When we pass that function reference by value to call_twice, it decays back to a function pointer.

Unary * can’t apply directly to a lambda such as two, so two implicitly converts to a function pointer so that we can * it. That gives a function reference. When we pass that function reference by value to call_twice, it decays to a function pointer.

The upshot is that call_twice is only ever instantiated once, with F = void(*)(). Godbolt:

int instantiations = 0;

template<class F>
void myTemplate(F *f) {
    static int x = ++instantiations;
    printf("Instantiation %d: ", x); (*f)();
}

int main() {
    auto a = []() { puts("A"); };
    auto b = []() { puts("B"); };
    myTemplate(&a);  // Instantiation 1: A
    myTemplate(&b);  // Instantiation 2: B
    myTemplate(*a);  // Instantiation 3: A
    myTemplate(*b);  // Instantiation 3: B
}

Anyway, I thought this was an interesting tidbit.

Posted 2020-03-31