Is it possible to use SFINAE and std::enable_if
to disable a single member function of a template class?
I currently have a code similar to this:
#include <type_traits>
#include <iostream>
#include <cassert>
#include <string>
class Base {
public:
virtual int f() { return 0; }
};
template<typename T>
class Derived : public Base {
private:
T getValue_() { return T(); }
public:
int f() override {
assert((std::is_same<T, int>::value));
T val = getValue_();
//return val; --> not possible if T not convertible to int
return *reinterpret_cast<int*>(&val);
}
};
template<typename T>
class MoreDerived : public Derived<T> {
public:
int f() override { return 2; }
};
int main() {
Derived<int> i;
MoreDerived<std::string> f;
std::cout << f.f() << " " << i.f() << std::endl;
}
Ideally, Derived<T>::f()
should be disabled if T != int
. Because f
is virtual, Derived<T>::f()
gets generated for any instantiation of Derived
, even if it is never called.
But the code is used such that Derived<T>
(with T != int
) never gets created only as a base class of MoreDerived<T>
.
So the hack in Derived<T>::f()
is necessary to make the program compile; the reinterpret_cast
line never gets executed.
You could simply specialize f
for int
:
template<typename T>
class Derived : public Base {
private:
T getValue_() { return T(); }
public:
int f() override {
return Base::f();
}
};
template <>
int Derived<int>::f () {
return getValue_();
}
No you can't rule out a member function with SFINAE. You could do it with specialisation of your Derived
class f
member function for convertible T
s to int
but that would lead to unnecessary duplication of code. In C++17 however you could solve this with use of if constexpr
:
template<typename T> class Derived : public Base {
T getValue_() { return T(); }
public:
int f() override {
if constexpr(std::is_convertible<T, int>::value) return getValue_();
return Base::f();
}
};
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