混合模板函数重载和继承

Question

以下代码打印:

generic
overload

但我想要的是在两种情况下都调用了重载或特化,而不是通用的.我不是试图将重载与模板专业化混合在一起,因为它们没有像我预期的那样工作.是否有任何模板魔法来实现这一目标？

#include <iostream>

class Interface {};
class Impl: public Interface {};

class Bar
{
public:
    template<typename T> void foo(T& t) {
        std::cout << "generic\n";
    }
    void foo(Interface& t) {
        std::cout << "overload\n";
    }
};
template<> void Bar::foo<Interface>(Interface& t) {
    std::cout << "specialization\n";
}

int main() {
    Bar bar;
    Impl impl;
    Interface& interface = impl;
    bar.foo(impl);
    bar.foo(interface);
    return 0;
}

Answer 1

type_traits用于测试参数是否派生自Interface的两种方法.

#include <boost/type_traits.hpp>

class Interface {};
class Impl: public Interface {};

class Bar
{
    template <class T> void foo_impl(T& value, boost::false_type)
    {
        std::cout << "generic\n";
    }
    void foo_impl(Interface& value, boost::true_type)
    {
        std::cout << "Interface\n";
    }
public:
    template<typename T> void foo(T& t) {
        foo_impl(t, boost::is_base_of<Interface, T>());
    }

};

如果满足条件,则禁用模板,仅将非模板作为候选模板.

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

class Interface {};
class Impl: public Interface {};

class Bar
{
public:
    template<typename T>
    typename boost::disable_if<boost::is_base_of<Interface, T>, void>::type foo(T& t)
    {
        std::cout << "generic\n";
    }

    void foo(Interface&)
    {
        std::cout << "Interface\n";
    }
};