When does type information flow backwards in C++?

Question

I just watched Stephan T. Lavavej talk at CppCon 2018 on \"Class Template Argument Deduction\", where at some point he incidentally says:

In

Answer 1

Here is at least one case:

struct foo {
  template<class T>
  operator T() const {
    std::cout << sizeof(T) << "\n";
    return {};
  }
};

if you do foo f; int x = f; double y = f;, type information will flow "backwards" to figure out what T is in operator T.

You can use this in a more advanced way:

template<class T>
struct tag_t {using type=T;};

template<class F>
struct deduce_return_t {
  F f;
  template<class T>
  operator T()&&{ return std::forward<F>(f)(tag_t<T>{}); }
};
template<class F>
deduce_return_t(F&&)->deduce_return_t<F>;

template<class...Args>
auto construct_from( Args&&... args ) {
  return deduce_return_t{ [&](auto ret){
    using R=typename decltype(ret)::type;
    return R{ std::forward<Args>(args)... };
  }};
}

so now I can do

std::vector<int> v = construct_from( 1, 2, 3 );

and it works.

Of course, why not just do {1,2,3}? Well, {1,2,3} isn't an expression.

std::vector<std::vector<int>> v;
v.emplace_back( construct_from(1,2,3) );

which, admittedly, require a bit more wizardry: Live example. (I have to make the deduce return do a SFINAE check of F, then make the F be SFINAE friendly, and I have to block std::initializer_list in deduce_return_t operator T.)

Answer 2

Stephan T. Lavavej explained the case he was talking about in a tweet:

The case I was thinking of is where you can take the address of an overloaded/templated function and if it’s being used to initialize a variable of a specific type, that will disambiguate which one you want. (There’s a list of what disambiguates.)

we can see examples of this from cppreference page on Address of overloaded function, I have excepted a few below:

int f(int) { return 1; } 
int f(double) { return 2; }   

void g( int(&f1)(int), int(*f2)(double) ) {}

int main(){
    g(f, f); // selects int f(int) for the 1st argument
             // and int f(double) for the second

     auto foo = []() -> int (*)(int) {
        return f; // selects int f(int)
    }; 

    auto p = static_cast<int(*)(int)>(f); // selects int f(int)
}

Michael Park adds:

It's not limited to initializing a concrete type, either. It could also infer just from the number of arguments

and provides this live example:

void overload(int, int) {}
void overload(int, int, int) {}

template <typename T1, typename T2,
          typename A1, typename A2>
void f(void (*)(T1, T2), A1&&, A2&&) {}

template <typename T1, typename T2, typename T3,
          typename A1, typename A2, typename A3>
void f(void (*)(T1, T2, T3), A1&&, A2&&, A3&&) {}

int main () {
  f(&overload, 1, 2);
}

which I elaborate a little more here.

Answer 3

I believe in static casting of overloaded functions the flow goes the opposite direction as in usual overload resolution. So one of those is backwards, I guess.