CODEWITHSUNDEEP
c++boostoperator-keywordsfinaeenable-if
Andrew Guo
Andrew Guo

Reputation: 13

Couldn't match the template method using boost enable_if

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

using namespace boost;

struct Serializable{};

struct Derived : Serializable {};

class Serializer
{
public:
    template <typename ValueType>
    void operator& (const typename enable_if<is_base_of<Serializable, ValueType>, ValueType>::type& value) {
        ;
    }
};

int main() {
    Serializer serializer;
    Derived tar;

    serializer & tar;

    return 0;
}

I was using g++4.4.7 which gave me a compile error: error: no match for ‘operator&’ in ‘serializer & tar’

How to make the specialization happen?

Upvotes: 1

Views: 68

Answers (1)

Igor R.
Igor R.

Reputation: 15075

The right-side value of the operator & should be of type ValueType. So, make the enable_if part a (fake) result type:

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

using namespace boost;

struct Serializable{};

struct Derived : Serializable {};

class Serializer
{
public:
    template <typename ValueType>
    const typename enable_if<is_base_of<Serializable, ValueType>, ValueType>::type* operator& (ValueType &v)
    {
      return 0;
    }
};

int main() 
{
    Serializer serializer;
    Derived tar;
    int i = 1;

    serializer & tar; //ok
    serializer & i; //error

    return 0;
}

In C++11 you could also make the enable_if construct a default template parameter, which would leave the function signature clean and unmodified:

template <typename ValueType, typename = typename enable_if<is_base_of<Serializable, ValueType>, ValueType>::type>
void operator& (ValueType &v)
{
}

Upvotes: 1

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