Question

我想通过模板参数在编译时切换默认构造函数的定义。我可以让它为转换构造函数编译OK，但尝试使用该方法默认构造函数是否默认 - 如果在特定模板参数的情况下，生成的类可能是POD，但在另一种情况下，它不能 - 但这样做时我得到编译器错误。如果没有专门化模板并复制所有相同的代码，有没有办法做到这一点？这是我正在尝试的简化版本：

#include<type_traits>   // for enable_if

template <bool MyParameter>
class Demonstration
{
    public:

        //trivial copy, move constructors/assignment, and trivial destructor
        constexpr Demonstration(Demonstration const &) = default;
        constexpr Demonstration(Demonstration &&) = default;
        Demonstration & operator= (Demonstration const &) = default;
        Demonstration & operator= (Demonstration &&) = default;
        ~Demonstration() = default;

        // this one gives "error: a template cannot be defauled"
        template <bool Dummy=MyParameter, typename std::enable_if< Dummy , bool >::type=true >
        Demonstration() = default;

        // ok
        template <bool Dummy=MyParameter, typename std::enable_if< !Dummy , bool >::type=false >
        Demonstration() : myValue(0) {}

        // ok
        template <bool Dummy=MyParameter, typename std::enable_if< Dummy , bool >::type=true >
        explicit constexpr Demonstration(unsigned char toConvert)
        : myValue ( toConvert )
        {
        }
        // ok
        template <bool Dummy=MyParameter, typename std::enable_if< !Dummy , bool >::type=false >
        explicit constexpr Demonstration(unsigned char toConvert)
        : myValue ( toConvert > 100 ? 0 : toConvert )
        {
        }

    // a lot of functions that do not depend on parameter go here

    protected:
    private:
        unsigned char myValue;

};

Answer 1

GCC抱怨你的模板：

error: a template cannot be defaulted

和Clang抱怨道：

error: only special member functions may be defaulted.

这看起来很公平。成员函数模板不是成员函数，更别说一个特别的了。

当Demonstration<bool P>为真时，您希望P成为POD 不一定如此。

一种可能的解决方案是完全委托POD-ness的参数化到基础模板base<bool P>的特化和拥有 Demonstration<P>继承base<P>。这是一个例子：

#include<type_traits>

template<bool Param = true>
struct base // is POD 
{
    base() = default;
    explicit constexpr base(unsigned char ch)
    : _val(ch){}
    unsigned char _val;
};

template<>
struct base<false> // is not POD
{
    base() = default;
    explicit constexpr base(unsigned char ch)
    : _val(ch > 100 ? 0 : ch){}
    unsigned char _val = 0;
};


template <bool MyParameter>
class Demonstration : private base<MyParameter>
{   
public:

    Demonstration() = default;
    //trivial copy, move constructors/assignment, and trivial destructor
    constexpr Demonstration(Demonstration const &) = default;
    constexpr Demonstration(Demonstration &&) = default;
    Demonstration & operator= (Demonstration const &) = default;
    Demonstration & operator= (Demonstration &&) = default;
    ~Demonstration() = default;

    explicit constexpr Demonstration(unsigned char toConvert)
    : base<MyParameter>(toConvert)
    {
    }

    char myValue() const {
        return base<MyParameter>::_val;
    }
};


#include <iostream>

using namespace std;

int main()
{
    cout << "is_pod<base<true>>::value = " 
        << is_pod<Demonstration<true>>::value << endl;
    cout << "is_pod<base<false>>::value = " 
        << is_pod<Demonstration<false>>::value << endl;
    cout << "is_pod<Demonstration<true>>::value = " 
        << is_pod<Demonstration<true>>::value << endl;
    cout << "is_pod<Demonstration<false>>::value = " 
        << is_pod<Demonstration<false>>::value << endl;
    Demonstration<true> d_true(1);
    Demonstration<false> d_false(101);
    std::cout << "(int)Demonstration<true>(1).myValue() = " 
        << (int)d_true.myValue() << endl;
    std::cout << "(int)Demonstration<false>(101).myValue() = " 
        << (int)d_false.myValue() << endl;
    return 0;
}

现在Demonstration<P>是POD，以防base<P>是POD。该程序输出：

is_pod<base<true>>::value = 1
is_pod<base<false>>::value = 0
is_pod<Demonstration<true>>::value = 1
is_pod<Demonstration<false>>::value = 0
(int)Demonstration<true>(1).myValue() = 1
(int)Demonstration<false>(101).myValue() = 0

使用GCC 4.8.2和clang 3.3构建

模板化默认构造函数

1 个答案: