如何在类模板中使用boost :: enable_if

Question

I am trying to use the boost::enable_if to turn on/off some functions in the class template but always get the compilation error error: no type named "type" in struct boost::enable_if .

My snippet:

#include <iostream>
#include <tr1/type_traits>
#include <boost/utility.hpp>

namespace std {
    using namespace tr1;
}

template <typename T1>
struct C {
    template< typename T2 >
    void test( T2&, typename boost::enable_if<
    std::is_const< T1 >, T1 >::type* = 0 ) {

        std::cout << "const" << std::endl;
    }

    template< typename T2 >
    void test( T2&, typename boost::disable_if<
    std::is_const< T1 >, T1 >::type* = 0 ) {

        std::cout << "non-const" << std::endl;
    }
};

int main() {
    const int ci = 5;
    int i = 6;

    C<char> c;
    c.test(ci);
    c.test(i);
    return 0;
}

But the following similar codes work fine:

#include <iostream>
#include <tr1/type_traits>
#include <boost/utility.hpp>

namespace std {
    using namespace tr1;
}

template <typename T1>
struct C {
    template< typename T2 >
    void test( T2&, typename boost::enable_if<
    std::is_const< T2 >, T1 >::type* = 0 ) {

        std::cout << "const" << std::endl;
    }

    template< typename T2 >
    void test( T2&, typename boost::disable_if<
    std::is_const< T2 >, T1 >::type* = 0 ) {

        std::cout << "non-const" << std::endl;
    }
};

int main() {
    const int ci = 5;
    int i = 6;

    C<char> c;
    c.test(ci);
    c.test(i);
    return 0;
}

What I want to achieve is to disable/enable some member functions based the types declared in the class template. Actually the template member function is not needed. They're only added for SFINAE.

Anyone can help??

Thanks!

Answer 1

SFINAE (which the mechanism used to implement enable_if ) only works in context of the function template's template parameters. In your case, T1 is a template parameter of the enclosing class template, not of the function template itself. From the function template's point of view, it's a fixed type and not being able to use it the way it's spelled out in the declaration is a normal error, not a substitution failure.

Answer 2

One way would be a specialization of the class itself, possibly as a base class in case you only want to do this for some functions:

template <typename T1>
struct B {
    template<typename T2>
    void test( T2& ) {

        std::cout << "non-const" << std::endl;
    }
};

template <typename T1>
struct B< const T1 > {
    template<typename T2>
    void test( T2& ) {

        std::cout << "const" << std::endl;
    }
};

template <typename T1>
struct C : B<T1> {
    //...
};