Pfthb

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

In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few.

Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question:

In which cases the C++17 standard mandates that type information propagate backwards?

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.)

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.

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.