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How can I enforce two function parameters have the same template type?


How can I enforce two function parameters have the same template type?

By : VDBK
Date : October 22 2020, 06:10 PM
To fix this issue You can check the data type at compile time, if you create the pointer inside your function call:
code :
template <typename TSensor, typename TProcessing, typename = std::enable_if_t<std::is_base_of_v<Sensor, TSensor> && std::is_base_of_v<SensorProcessing, TProcessing>>>
void register_sensor(std::string const& name, TSensor s, TProcessing p)
{
    // enforce that p eats what s produces, in a typesafe but polymorphic manner
    static_assert(std::is_same_v<typename TSensor::data_type, typename TProcessing::data_type>, "Data type missmatch");

    // create the pointers here and move s and p in
}


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Enforce template function parameters to be iterators on a specific type

Enforce template function parameters to be iterators on a specific type


By : John Daniel
Date : March 29 2020, 07:55 AM
Any of those help The question is little vague and has a few issues. But I think I get the gist of it. The static_assert you are looking for probably;
code :
static_assert(std::is_base_of<A,
  typename std::iterator_traits<IT1>::value_type>::value,
  "wrong type");
Is there a good way to enforce type restrictions on function parameters in a variadic template in C++?

Is there a good way to enforce type restrictions on function parameters in a variadic template in C++?


By : Andrew
Date : March 29 2020, 07:55 AM
This might help you One easy way to do it would be to create a mapping from the enumerators to the desired types and use that to build the function parameter list - you could think of it as "enumerator traits", I guess:
code :
#include <iostream>
#include <string>

enum Type {STRING, TYPE_A_INT, TYPE_B_INT};

template<Type> struct type_from;

template<> struct type_from<STRING> { using type = std::string; };
template<> struct type_from<TYPE_A_INT> { using type = int; };
template<> struct type_from<TYPE_B_INT> { using type = int; };

template<Type E> using type_from_t = typename type_from<E>::type;

template<Type... Es> void Foo(type_from_t<Es>... args)
{
   // Do stuff with args.
   using expander = int[];
   (void)expander{0, (std::cout << args << ' ', 0)...};
   std::cout << '\n';
}

int main()
{
   Foo<STRING, TYPE_A_INT>("str", 32); // works
   Foo<STRING, TYPE_B_INT>("str", 32);  // works
   Foo<STRING, TYPE_B_INT, TYPE_A_INT, STRING>("str", 32, 28, "str");  // works
   // Foo<STRING, TYPE_B_INT>("str", "str");  // doesn't work
}
Call template function with non-type parameters explicit and type parameters implicit

Call template function with non-type parameters explicit and type parameters implicit


By : SUKLA DUTTA BANIK
Date : March 29 2020, 07:55 AM
it helps some times You can, but deduced template parameters need to be at the end of the argument list. You can make your code compile by reordering parameters of your function template:
code :
template < bool bPrint=true, typename T>
void f(T var) {
  if (bPrint)
    std::cout <<  typeid(var).name() << std::endl;
}
Enforce the return type of a function based on a property in its parameters

Enforce the return type of a function based on a property in its parameters


By : user1956141
Date : March 29 2020, 07:55 AM
I wish this helpful for you TypeScript won't narrow type parameters via type guards. That means the check action.type === "type1" does narrow action.type, but it doesn't narrow T, so the return type is still something like the union type void | {foo: "bar"}. It's apparently not an easy problem to solve, as @RyanCavanaugh said:
code :
type Ret<T extends GameActionTypes> = 
  FindByTag<ActionType<gameActions>, {type: T}>["returnType"];

export function executeAction<T extends GameActionTypes>(
  action: ActionType<gameActions>
): Ret<T> {
  if (action.type === "type1") {
    type R = Ret<typeof action.type>;
    return undefined as R; // okay
  } else if (action.type === "type2") {
    type R = Ret<typeof action.type>;
    return undefined as R; // error
  }
}
export function executeAction<A extends ActionType<gameActions>>(
  action: A
): A["returnType"] {
  const actionUnion: ActionType<gameActions> = action; // remove generic
  if (actionUnion.type === "type1") {
    type R =  Ret<typeof actionUnion.type>
    return undefined as R;
  } else if (action.type === "type2") {
    type R =  Ret<typeof actionUnion.type>
    return undefined as R;
  }
}
declare const t1: TEST1;
const ret1 = executeAction(t1); // void
declare const t2: TEST2;
const ret2 = executeAction(t2); // {foo: "bar"}
Generic constraint: Enforce type to have static function and constructor with parameters

Generic constraint: Enforce type to have static function and constructor with parameters


By : Jannis Relakis
Date : March 29 2020, 07:55 AM
This might help you No, there's nothing like this in C#.
I've previously suggested that "static interfaces" could express this reasonably neatly. They'd only be useful for generic type constraints (I suspect, anyway) but then you could express:
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