C 17如果用元组用constexpr()使用

C++17 if constexpr() usage with tuple

本文关键字:使用 constexpr 元组 如果      更新时间:2023-10-16

我正在使用启用C 17支持的VS2017。

我想尝试制作一个"变压器"类,当提供某个类型的类型时,它将转换类型,否则它将像原样返回变量。目标是将所有变量类型传递给变压器,并"隐藏"其变换。这样,呼叫者可以尝试改变一切,而不必担心变换是否需要,变压器会知道。

一个更完整的示例(从原始编辑):

    class MyPoint
{
public:
    int x = 0;
};
class NotMyPoint
{
public:
    int x = 50;
};
template <typename T>
class ITransform
{
    public:
    virtual ~ITransform() {};
    virtual T InTransform(const T &in) const = 0;
    virtual T OutTransform(const T &out) const = 0;
    //Check if the argument type is the same as this class type
    template <typename X>
    constexpr bool CanTransform() const
    {
        return std::is_same<X, T>::value;
    }
};
class MyTransformer :
    public ITransform<MyPoint>
{
public:
    MyTransformer() = default;
    virtual MyPoint InTransform(const MyPoint &in) const override
    {
        auto newPt = in;
        newPt.x += 100;
        return newPt;
    }
    virtual MyPoint OutTransform(const MyPoint &in) const override
    {
        auto newPt = in;
        newPt.x -= 100;
        return newPt;
    }
};
template <class... TRANSFORMERS>
struct VariadicTransformer
{
    constexpr VariadicTransformer() = default;
    /** brief parse using validateParse but catch throw */
    template <typename T>
    inline T Transform(const T& in)
    {
        return TransformImpl<sizeof...(TRANSFORMERS)-1, T>(in);
    }
private:
    /// last attempt to find matching transformer at I==0, if it fails return the original value
    template<std::size_t I = 0, typename T>
    inline typename std::enable_if<I == 0, T>::type TransformImpl(const T &in) const
    {
        if (std::get<I>(transformers).CanTransform<T>())
            return std::get<I>(transformers).InTransform(in);
        else
            return in;
    }
    /// attempt to find transformer for this type
    template<std::size_t I = 0, typename T>
    inline typename std::enable_if < I != 0, T>::type TransformImpl(const T &in) const
    {
        if (std::get<I>(transformers).CanTransform<T>())
            return std::get<I>(transformers).InTransform(in);
        else
            return TransformImpl<I - 1, T>(in);
    }
    std::tuple<const TRANSFORMERS...> transformers;
};

//Example usage
VariadicTransformer<MyTransformer, MyTransformer> varTrans;
MyPoint myPoint;
NotMyPoint notMyPoint;
std::cout << myPoint.x << std::endl;
myPoint = varTrans.Transform(myPoint);
std::cout << myPoint.x << std::endl;
std::cout << notMyPoint.x << std::endl;
notMyPoint = varTrans.Transform<NotMyPoint>(notMyPoint);
std::cout << notMyPoint.x << std::endl;
return 0;

我的问题带有这一行:

if constexpr(std::get<I>(transformers).CanTransform<T>())

这不会编译并提供以下错误:

错误C2131:表达未评估为常数

注意:失败是由于其寿命以外的变量读取

引起的

注意:请参阅"此'''

的用法

cantransform函数应该是constexpr,std :: get&lt;#>(std :: tuple)应该是constexpr,所以我不确定其对这一行的投诉是什么。

也需要if constexpr来避免试图调用任何不符合转换当前类型的变压器,我希望这种情况掉落并返回原始类型。

关于导致此错误或我可以尝试的其他设计的任何建议?

一个方法调用仅当调用对象为 constexpr时,只有constexpr。如果调用对象不是constexpr,则该方法仍将在运行时而不是编译时进行评估,因此不符合任何汇编评估的条件。

struct A {
    int val;
    constexpr A() : A(16) {}
    constexpr A(int val) : val(val) {}
    constexpr bool foo() const {return val > 15;}
};
int main() {
    A a;
    if constexpr(a.foo()) {
        std::cout << "No point thinking about it; this won't compile!" << std::endl;
    } else {
        std::cout << "Again, the previous line doesn't compile." << std::endl;
    }
    constexpr A b;
    if constexpr(b.foo()) {
        std::cout << "This, however, will compile, and this message will be displayed!" << std::endl;
    }
    constexpr A c(13);
    if constexpr(c.foo()) {
        std::cout << "This will not be displayed because the function will evaluate to false, but it will compile!" << std::endl;
    }
}

您需要确保可以制作TransformImpl constexpr,然后确保A的实例TransformImpl也为constexpr

,所以我无法解释为什么上一个选项不起作用。但是,但是,通过将CanTransForm函数变成其自己的独立函数,错误就消失了。

这最终有效:

template<typename X, typename T>
constexpr inline bool CanTransform()
{
    return std::is_base_of<ITransform<T>, X>::value;
}
template <class... TRANSFORMERS>
struct VariadicTransformer
{
    constexpr VariadicTransformer() = default;
    template <typename T>
    constexpr inline T Transform(const T& in) const
    {
        return TransformImpl<sizeof...(TRANSFORMERS)-1, T>(in);
    }
private:
        // last attempt to find matching transformer at I==0, if it fails return the original value
    template<std::size_t I = 0, typename T>
    constexpr inline typename std::enable_if<I == 0, T>::type TransformImpl(const T &in) const
    {
        if constexpr(CanTransform < std::tuple_element < I, std::tuple<TRANSFORMERS...>>::type, T >())
            return std::get<I>(transformers).InTransform(in);
        else
            return in;
    }
       // attempt to find transformer for this type
    template<std::size_t I = 0, typename T>
    constexpr inline typename std::enable_if < I != 0, T>::type TransformImpl(const T &in) const
    {
        if constexpr(CanTransform < std::tuple_element < I, std::tuple<TRANSFORMERS...>>::type, T >())
            return std::get<I>(transformers).InTransform(in);
        else
            return TransformImpl<I - 1, T>(in);
    }
    std::tuple<TRANSFORMERS...> transformers;
};