模糊元函数或未定义类型-Java 学习之路

我是metafunctions的新手 . 我想编写一个函数，用一些其他类型替换复合类型中某种类型的所有匹配项 . 例如： replace<void *, void, int>::type 应为 int * ， replace<void, void, int>::type 应为 int 等 .

到目前为止我基本上失败了两种不同的方法：

template
    <
        typename C, // Type to be searched
        typename X, // "Needle" that is searched for
        typename Y  // Replacing type
    >
struct replace
{
    typedef C type;
};

// If the type matches the search exactly, replace
template
    <
        typename C,
        typename Y
    >
struct replace<C, C, Y>
{
    typedef Y type;
};

// If the type is a pointer, strip it and call recursively
template
    <
        typename C,
        typename X,
        typename Y
    >
struct replace<C *, X, Y>
{
    typedef typename replace<C, X, Y>::type * type;
};

这对我来说似乎很简单，但我发现当我尝试 replace<void *, void *, int> 时，编译器无法决定是否在这种情况下使用 replace<C, C, Y> 或 replace<C *, X, Y> ，因此编译失败 .

我试过的下一件事就是在基函数中剥离指针：

template
    <
        typename C,
        typename X,
        typename Y
    >
struct replace
{
    typedef typename boost::conditional
        <
            boost::is_pointer<C>::value,
            typename replace
                <
                    typename boost::remove_pointer<C>::type,
                    X, Y
                >::type *,
            C
        >::type
    type;
};

......这时我发现我也不能这样做，因为 type 显然没有在那时定义，所以我不能从基函数做递归 typedef .

现在我没有想法 . 你怎么解决这个问题？

3 回答

你试过的东西：

#include <typeinfo>
#include <type_traits>

template<typename C, typename X, typename Y>
struct replace {
private:

    typedef
        typename std::conditional <
            std::is_pointer<C>::value,
            typename std::remove_pointer<C>::type,
            C >::type
        strippedT;

    typedef
        typename std::conditional <
            std::is_same<strippedT, X>::value,
            Y,
            strippedT>::type
        replacedT;
public:

    typedef
        typename std::conditional <
            std::is_pointer<C>::value,
            replacedT*,
            replacedT >::type
        type;
};

int main()
{
    typedef replace<void*, void, int>::type T1;
    std::cout << typeid(T1).name() << '\n';  // prints Pi on my implementation

    typedef replace<void, void, int>::type T2;
    std::cout << typeid(T2).name();          // prints i
}

Kerrek的答案看起来好多了:)

回复于 2024-05-08T09:31:02+08:00

这是一个大致的想法：

template <typename, typename> struct pattern;

template <typename T> struct pattern<T, T>
{
    template <typename U> struct rebind
    {
        typedef U other;
    };
};

template <typename A, typename B> struct pattern<A*, B>
{
    template <typename U> struct rebind
    {
        typedef typename pattern<A, B>::template rebind<U>::other * other;
    };
};

template <typename Haystack, typename Needle, typename New>
struct replace
{
    typedef typename pattern<Haystack, Needle>::template rebind<New>::other type;
};

测试：

#include <demangle.hpp>
#include <iostream>
int main()
{
    typedef replace<void, void, int>::type T1;
    typedef replace<void*, void, int>::type T2;

    std::cout << demangle<T1>() << std::endl;
    std::cout << demangle<T2>() << std::endl;
}

打印：

int
int*

Edit: 这是一个更完整的集合：

template <typename, typename> struct pattern;
template <typename, typename> struct pattern_aux;

template <typename A, typename B> struct pattern_aux
{
    template <typename U> struct rebind
    {
        typedef typename pattern<A, B>::template rebind<U>::other other;
    };
};

template <typename A, typename B, unsigned int N> struct pattern_aux<A[N], B>
{
    template <typename U> struct rebind
    {
        typedef typename pattern<A, B>::template rebind<U>::other other[N];
    };
};


template <typename A, typename B> struct pattern
{
    template <typename U> struct rebind
    {
        typedef typename pattern_aux<A, B>::template rebind<U>::other * other;
    };
};

template <typename T> struct pattern<T, T>
{
    template <typename U> struct rebind
    {
        typedef U other;
    };
};

template <typename A, typename B> struct pattern<A*, B>
{
    template <typename U> struct rebind
    {
        typedef typename pattern<A, B>::template rebind<U>::other * other;
    };
};

template <typename A, typename B> struct pattern<A const, B>
{
    template <typename U> struct rebind
    {
        typedef typename pattern_aux<A, B>::template rebind<U>::other const other;
    };
};

template <typename A, typename B> struct pattern<A volatile, B>
{
    template <typename U> struct rebind
    {
        typedef typename pattern_aux<A, B>::template rebind<U>::other volatile other;
    };
};

template <typename A, typename B> struct pattern<A const volatile, B>
{
    template <typename U> struct rebind
    {
        typedef typename pattern_aux<A, B>::template rebind<U>::other const volatile other;
    };
};

template <typename Haystack, typename Needle, typename New>
struct replace
{
    typedef typename pattern<Haystack, Needle>::template rebind<New>::other type;
};

回复于 2024-05-08T09:31:02+08:00

按照您的代码，为什么不再添加一个专业化

template
    <
        typename C,
        typename Y
    >
struct replace<C*, C, Y>
{
    typedef Y* type;
};

实时代码example results

或者更多：

template<class T>
struct tag_t {using type=T;};

template <class C, class X, class Y>
struct replace {};

template <class C, class Y>
struct replace<C, C, Y>: tag_t<Y> {};

template <class C, class Y>
struct replace<C*, C, Y>: tag_t<Y*>{};

实时代码results

回复于 2024-05-08T09:31:02+08:00

模糊元函数或未定义类型

3 回答

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