I have the following bit of code that uses the "deduce this" pattern:
#include <print>
struct base
{
template <typename Self>
void invoke(this Self& self)
{
self.foo();
}
};
struct derived : base
{
void foo() { std::println("derived::foo()"); }
};
int main()
{
derived d;
d.invoke(); // ok
base& bref = d;
bref.invoke(); // error
base* bptr = &d;
bptr->invoke(); // error
return 0;
}
The compiler generates the following error:
<source>:8:12: error: no member named 'foo' in 'base'
8 | self.foo();
| ~~~~ ^
<source>:23:6: note: in instantiation of function template specialization 'base::invoke<base>' requested here
23 | bref.invoke(); // error
| ^
Given this error, is the base type used for the "deduced this" not the same kind of type one would expect when using normal runtime polymorphism?
Given the error generated by the compiler, is the base type used for the deduce this not the same kind of type one would expect when using normal runtime polymorphism?
Deduced this is compile-time polymorphism, just like templates in general.
For any given use of invoke in the program, the type of Self will be deduced (at compile-time) from the object expression of the call, the function foo will be looked up in that type and only that function can be called. There is no runtime polymorphism involved. For runtime polymorphism you need virtual functions.
When you call invoke on an expression of type base rather than derived, then Self and the type of self in invoke will also be base. Looking up foo in base will not find anything, because you never declared a foo function in it.
For runtime polymorphism you need to declare foo as virtual in base and then it will work as you seem to expect regardless of whether invoke is called on a base or derived type. However, in that case the function doesn't need to use deduced-this at all. A normal non-static member function will do just fine, as would calling foo directly, instead of going through invoke.