Okay, this is a bit complicated so please bear with me. :)
class A {};
class DA : public A {};
class DDA : public DA {};
void f(A x) {
std::cout << "f A" << std::endl;
}
void f(DA x) {
std::cout << "f DA" << std::endl;
}
void f(DDA x) {
std::cout << "f DDA" << std::endl;
}
Now we want to add another function that treats DA a little differently.
void g(A t) {
std::cout << "generic treatment of A" << std::endl;
std::cout << "called from g: ";
f(t);
}
void g(DA t) {
std::cout << "special treatment of DA" << std::endl;
std::cout << "called from g: ";
f(t);
}
But calling this with an object of each of the classes clearly does not have the desired effect.
Call:
A a; DA b; DDA c;
g(a); g(b); g(c)
Result:
generic treatment of A
called from g: f A
special treatment of DA
called from g: f DA
special treatment of DA
called from g: f DA //PROBLEM: g forgot that this DA was actually a DDA
template<typename T>
void h(T t) {
std::cout << "generic treatment of A" << std::endl;
std::cout << "called from h: ";
f(t);
}
template<>
void h<>(DA t) {
std::cout << "special treatment of DA" << std::endl;
std::cout << "called from h: ";
f(t);
}
which results in:
generic treatment of A
called from h: f A
special treatment of DA
called from h: f DA
generic treatment of A //PROBLEM: template specialization is not used
called from h: f DDA
Well, how about we don't use template specialization but define a non-template function for the special case? (Article on the very confusing matter.) It turns out it behaves in exactly the same way because the non-template function which is according to the article a "first class citizen", seems to lose because a type conversion is necessary to use it. And if it would be used, well then we would just be back at the first solution (I assume) and it would forget the type of DDA.
template<typename T>
void i(T t, void* magic) {
std::cout << "generic treatment of A" << std::endl;
std::cout << "called from i: ";
f(t);
}
template<typename T>
void i(T t, DA* magic) {
std::cout << "special treatment of DA" << std::endl;
std::cout << "called from i: ";
f(t);
}
But it seems to do exactly what I want:
generic treatment of A
called from i: f A
special treatment of DA
called from i: f DA
special treatment of DA
called from i: f DDA
Even though it needs to be called in a weird way: i(a, &a); i(b, &b); i(c, &c);
Now I have several questions:
I hope this was reasonably clear. :)
The function template overload
template<typename T>
void i(T t, DA* magic) {
will only be available if the parameter magic
is convertible to type DA *
. This is obviously the case for &b
but also for &c
as pointer-to-derived is convertible to pointer-to-base. The void *
function template overload is always available but DA *
is preferred to void *
, per §13.3.3.2:4:
c++11
13.3.3.2 Ranking implicit conversion sequences [over.ics.rank]
[...]
4 Standard conversion sequences are ordered by their ranks: an Exact Match is a better conversion than a Promotion, which is a better conversion than a Conversion. Two conversion sequences with the same rank are indistinguishable unless one of the following rules applies:
[...]
— If class
B
is derived directly or indirectly from classA
, conversion ofB*
toA*
is better than conversion ofB*
tovoid*
, and conversion ofA*
tovoid*
is better than conversion ofB*
tovoid*
.
As you've noted, it's a perfectly workable scheme; it would make more sense to wrap the magic in another template function that takes care of calling i
with (a, &a)
:
template<typename T>
void j(T t) {
i(t, &t);
}
In terms of safety, it's fine; if the DA *
overload is lost then the void *
overload will be selected silently; it's up to you to decide whether this is desirable.
As an alternative, you can use std::enable_if
to select between templates:
template<typename T>
typename std::enable_if<!std::is_base_of<DA, T>::value>::type g(T t) {
std::cout << "generic treatment of A" << std::endl;
f(t);
}
template<typename T>
typename std::enable_if<std::is_base_of<DA, T>::value>::type g(T t) {
std::cout << "special treatment of DA" << std::endl;
f(t);
}
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