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Approaches to function SFINAE in C++ [duplicate] (2 answers)

Why can I seemingly define a partial specialization for function templates? (7 answers)

Why can function templates not be partially specialized? (5 answers)

Closed 12 mins ago.

I m 试图根据使用<代码>的参数类型对功能模板进行专门化,类似于以下例子::https://en.cpp.com/w/cpp/types/enable_if

template<typename Type, typename Enabled = void>
void foo(Type t)
{
    std::cout << "primary
";
    std::cout << std::is_same<decltype(t), A>::value << "
";
}


template<typename Type, typename std::enable_if_t<std::is_same<Type, A>::value>::type>
void foo(Type t)
{
    std::cout << "specialized
";
}

然而,在这种情况下,只有主要定义才被要求。如果我试图与班级相同的话

template <class T, class Enable = void> 
struct Foo
{
    static void bar()
    {
        std::cout << "T is blah" << std::endl;
    }
};


template <class T> 
struct Foo<T, typename std::enable_if<std::is_integral<T>::value>::type>
{
    static void bar()
    {
        std::cout << "T is int" << std::endl;
    }
};

它实现了理想行为:

int main() {
    A a;

    // not specializing
    foo<A>(a); // primary called
    foo(19);   // primary called

    // does specializing
    Foo<int>::bar();  // int specialization is called
    Foo<float>::bar(); // generic is called

    return 0;
}

Could someone please explain this difference to me and if there is a way to achieve the desired behavior without use of classes?

My code example: https://godbolt.org/z/86jsxfqdG

Answer 1

这不是专业化,而是超负荷。功能模板不能像班级模板那样部分专门化。第2次<代码>foo在您的代码中获胜,因为其第2次模板参数可以推卸。

你们

// overload for types other than A
template<typename Type>
typename std::enable_if_t<!std::is_same<Type, A>::value> foo(Type t)
{
    std::cout << "overload for types other than A
";
}

// overload for type A
template<typename Type>
typename std::enable_if_t<std::is_same<Type, A>::value> foo(Type t)
{
    std::cout << "overload for type A
";
}

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