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Multiple SFINAE rules

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.

like image 666
Fred Avatar asked Apr 27 '10 02:04

Fred


2 Answers

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.

like image 176
James McNellis Avatar answered Sep 22 '22 19:09

James McNellis


The question is easy when you realize that

if (a) { X(); }
else if (b) { Y(); }

means exactly the same as

if (a) { X(); }
if (!a && b) { Y(); }

However, you could also extend your true/false dichotomy.

enum FuncVariants { HasMember, PointerDerivedFromA, DerivedFromB, InstanceOfC, isPod }
template<typename T>
void Func(T &arg, int_to_type<HasMember>);

template<typename T>
void Func(T &arg, int_to_type<DerivedFromA>);

template<typename T>
void Func(T &arg, int_to_type<DerivedFromB>);

template<typename T>
void Func(T &arg, int_to_type<InstanceOfC>);

(Obviously, when calling you have to take care as the options are not mutually exclusive)

like image 23
MSalters Avatar answered Sep 19 '22 19:09

MSalters