Multiple SFINAE rules

c++, sfinae, templates

Solution

This is certainly possible; you just have to be careful to ensure that all of the branches are mutually exclusive, otherwise you'll end up with an ambiguity.

Take a look at Boost Type Traits and Boost Enable If, which are the two best tools for supporting this. Boost ICE (which stands for Integral Constant Expression) can be used to combine multiple type traits to help you to do more complex type matching (and to ensure that your overloads are mutually exclusive.

This can be somewhat complicated and convoluted, so here's a relatively straightforward example. Say you have a class hierarchy:

struct Base { };
struct Derived : Base { };

and you want to call one overload of a function `foo` for `Base`, and another overload for any class derived from `Base`. A first attempt might look like:

#include <boost/type_traits.hpp>
#include <boost/utility/enable_if.hpp>

using namespace boost;
using namespace boost::type_traits;

template <typename T>
typename enable_if<is_same<Base, T>, void>::type
foo(const T&) { }

template <typename T>
typename enable_if<is_base_of<Base, T>, void>::type
foo(const T&) { }

However, `is_base_of` returns true if `T` is the base class, so if you attempt to call `foo(Base())`, there is an ambiguity because both function templates match. We can resolve this by using a combination of the type traits and using the Boost ICE helpers:

template <typename T>
typename enable_if<is_same<Base, T>, void>::type
foo(const T&) { }

template <typename T>
typename enable_if<
    ice_and<
        is_base_of<Base, T>::value,
        ice_not<is_same<Base, T>::value>::value 
    >, void>::type
foo(const T&) { }

These overloads are mutually exclusive, and they ensure there is no ambiguity.

Some of your examples are not supported (namely, `HAS_MEMBER_FUNCTION_X`; I'm not sure about `IS_TEMPLATE_CLASS_C`--depending on what you want to do with it you might be able to make something work), but in general this is possible.

Problem

After reading the answer to this question, I learned that SFINAE can be used to choose between two functions based on whether the class has a certain member function. It's the equivalent of the following, just that each branch in the if statement is split into an overloaded function: ``` template<typename T> void Func(T& arg) { if(HAS_MEMBER_FUNCTION_X(T)) arg.X(); else //Do something else because T doesn't have X() } ``` becomes ``` template<typename T> void Func(T &arg, int_to_type<true>); //T has X() template<typename T> void Func(T &arg, int_to_type<false>); //T does not have X() ``` I was wondering if it was possible to extend SFINAE to do multiple rules. Something that would be the equivalent of this: ``` template<typename T> void Func(T& arg) { if(HAS_MEMBER_FUNCTION_X(T)) //See if T has a member function X arg.X(); else if(POINTER_DERIVED_FROM_CLASS_A(T)) //See if T is a pointer to a class derived from class A arg->A_Function(); else if(DERIVED_FROM_CLASS_B(T)) //See if T derives from class B arg.B_Function(); else if(IS_TEMPLATE_CLASS_C(T)) //See if T is class C<U> where U could be anything arg.C_Function(); else if(IS_POD(T)) //See if T is a POD type //Do something with a POD type else //Do something else because none of the above rules apply } ``` Is something like this possible? Thank you.

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