4

我正在尝试调用这样的模板函数：

typedef std::tuple<int, double, bool> InstrumentTuple;

Cache cache;
InstrumentTuple tuple = cache.get<InstrumentTuple>();

我知道我可以“简单地”传递元组的类型。这是我所知道的，但这很麻烦，因为我对这个函数做了很多调用，而且元组很长：

InstrumentTuple tuple = c.get<int, double, bool>(); // syntax I'd like to avoid

所以我尝试了 get 方法的多种实现，但没有成功：

通过模板参数启用

#include <tuple>

class Cache
{
private:
    template<int I, typename T, typename = typename std::enable_if<I == std::tuple_size<T>::value>::type>
    std::tuple<> get() // line 6
    {
        return std::tuple<>();
    }

    template<int I, typename T, typename = typename std::enable_if<I != std::tuple_size<T>::value>::type>
    std::tuple<typename std::tuple_element<I,T>::type, decltype(get<I+1, T>())> get() // line 12
    {
        std::tuple<typename std::tuple_element<I,T>::type> value;
        return std::tuple_cat(value, get<I+1, T>());
    }

public:
    template<typename T>
    T get()
    {
        return get<0, T>(); // line 22
    }
};

int main(int argc, char** argv)
{
    Cache cache;
    typedef std::tuple<int, double, bool> InstrumentTuple;
    InstrumentTuple tuple = cache.get<InstrumentTuple>(); // line 30
}

这给了我这个错误：

main.cpp: In instantiation of 'T Cache::get() [with T = std::tuple<int, double, bool>]':
main.cpp:30:56:   required from here
main.cpp:22:26: error: no matching function for call to 'Cache::get()'
main.cpp:22:26: note: candidates are:
main.cpp:6:18: note: template<int I, class T, class> std::tuple<> Cache::get()
main.cpp:6:18: note:   template argument deduction/substitution failed:
main.cpp:5:33: error: no type named 'type' in 'struct std::enable_if<false, void>'
main.cpp:12:81: note: template<int I, class T, class> std::tuple<typename std::tuple_element<I, T>::type, decltype (get<(I + 1), T>())> Cache::get()
// ----- Important part
main.cpp:12:81: note:   template argument deduction/substitution failed:
main.cpp: In substitution of 'template<int I, class T, class> std::tuple<typename std::tuple_element<I, T>::type, decltype (get<(I + 1), T>())> Cache::get() [with int I = 0; T = std::tuple<int, double, bool>; <template-parameter-1-3> = <missing>]':
// -----
main.cpp:22:26:   required from 'T Cache::get() [with T = std::tuple<int, double, bool>]'
main.cpp:30:56:   required from here
main.cpp:12:81: error: no matching function for call to 'Cache::get()'
main.cpp:12:81: note: candidate is:
main.cpp:6:18: note: template<int I, class T, class> std::tuple<> Cache::get()
main.cpp:6:18: note:   template argument deduction/substitution failed:
main.cpp:5:33: error: no type named 'type' in 'struct std::enable_if<false, void>'
main.cpp: In instantiation of 'T Cache::get() [with T = std::tuple<int, double, bool>]':
main.cpp:30:56:   required from here
main.cpp:20:7: note: template<class T> T Cache::get()
main.cpp:20:7: note:   template argument deduction/substitution failed:
main.cpp:22:26: error: wrong number of template arguments (2, should be 1)

我不明白为什么缺少模板参数。

所以我尝试了另一种实现：

模板模板命名参数

#include <tuple>

class Cache
{
private:
    template<int>
    std::tuple<> get() // line 7
    {
        return std::tuple<>();
    }

    template<int index, typename type, typename... rest>
    std::tuple<type, rest...> get() // line 13
    {
        return std::tuple_cat(std::tuple<type>(), get<index+1, rest...>());
    }

public:
    template<template<typename... types> class tuple>
    typename std::tuple<(tuple::types)...> get()
    {
        return get<0, (tuple::types)...>();
    }
};  // line 24

int main(int argc, char** argv)
{
    Cache cache;
    typedef std::tuple<int, double, bool> InstrumentTuple;
    InstrumentTuple tuple = cache.get<InstrumentTuple>(); // line 30
}

但后来我得到这个错误：

// ----- Important part
main.cpp:24:1: error: expected identifier before '}' token
main.cpp:24:1: error: expected unqualified-id before '}' token
// -----
main.cpp: In function 'int main(int, char**)':
main.cpp:30:56: error: no matching function for call to 'Cache::get()'
main.cpp:30:56: note: candidates are:
main.cpp:7:18: note: template<int <anonymous> > std::tuple<> Cache::get()
main.cpp:7:18: note:   template argument deduction/substitution failed:
main.cpp:13:31: note: template<int index, class type, class ... rest> std::tuple<_Head, _Tail ...> Cache::get()
main.cpp:13:31: note:   template argument deduction/substitution failed:

同样，由于缺少标识符，我不明白错误。

我现在想知道我想要实现的目标是否可能。是否可以使用std::tuple我想要的？或者，还有更好的方法 ？

4

2 回答 2

6

您的第一个解决方案失败了，因为第二个重载 toget在其自己的返回类型声明时不可见；为了解决这个问题，您需要将返回类型计算分离到它自己的子程序中。

第二种解决方案更接近；问题是你只是在推断template std::tuple，而不是它的论点。推断可变参数（例如类型参数tuple）的一种简单方法是通过空标记结构，需要一个额外的间接级别：

template<typename T> struct type_tag {};

class Cache {
    // ... (as above)

    template<typename... Ts> std::tuple<Ts...> get(type_tag<std::tuple<Ts...>>) {
        return get<0, Ts...>();
    }

public:
    template<typename T> T get() {
        return get(type_tag<T>{});
    }
};

您应该检查是否可以使用包扩展而不是递归来编写解决方案，例如：

template<typename T> struct type_tag {};

class Cache {
    template<typename... Ts> std::tuple<Ts...> get(type_tag<std::tuple<Ts...>>) {
        return std::tuple<Ts...>{Ts{}...};
    }

public:
    template<typename T> T get() {
        return get(type_tag<T>{});
    }
};
于 2013-03-11T12:43:18.647 回答
3

这是我想出的：

#include <tuple>

struct Cache;

/* typename = std::tuple<...> */
template<int, typename> struct cache_getter;
/* typename = parameters from std::tuple<...> */
template<int, typename...> struct tuple_walker;

template<int I, typename... Ts> struct cache_getter<I, std::tuple<Ts...> > {
    static std::tuple<Ts...> get(Cache & c);
};

struct Cache {
protected:
    template<int, typename...> friend struct tuple_walker;
private:
    /* here T is a type from within a std::tuple<...> */
    template<int I, typename T> std::tuple<T> get_ex() {
        return std::tuple<T>();
    }
public:
    /* here T is actually a std::tuple<...> */
    template<typename T> T get() {
        return cache_getter<0, T>::get(*this);
    }
};

/* since std::tuple_cat only accepts 2 std::tuples per call but we don't have control over the number of types in the passed in std::tuple, we'll need to chain our calls */
template<typename...> struct my_tuple_cat;
template<typename H, typename... T> struct my_tuple_cat<H, T...> {
    static auto cat(H h, T... t) -> decltype(std::tuple_cat(h, my_tuple_cat<T...>::cat(t...)))
    { return std::tuple_cat(h, my_tuple_cat<T...>::cat(t...)); }
};
template<typename T> struct my_tuple_cat<T> {
    static T cat(T t) { return t; }
};

/* this one is used to call Cache.get_ex<int I, typename T>() with incrementing values for I */
template<int I, typename H, typename... T> struct tuple_walker<I, H, T...> {
    static std::tuple<H, T...> get(Cache & c) {
        return my_tuple_cat<std::tuple<H>, std::tuple<T...>>::cat(c.get_ex<I, H>(), tuple_walker<I + 1, T...>::get(c));
    }
};
template<int I, typename H> struct tuple_walker<I, H> {
    static std::tuple<H> get(Cache & c) {
        return c.get_ex<I, H>();
    }
};
/* this one will forward the types in std::tuple<...> to tuple_walker to get each tuple separately */
template<int I, typename... Ts> std::tuple<Ts...> cache_getter<I, std::tuple<Ts...> >::get(Cache & c) {
    return tuple_walker<I, Ts...>::get(c);
}

int main(int argc, char ** argv) {
    Cache cache;
    typedef std::tuple<int, double, bool> InstrumentTuple;
    InstrumentTuple tuple = cache.get<InstrumentTuple>();
    return 0;
}

我希望这是值得的。我还没有在 C++11 中做太多事情，所以也许这不是一个最佳解决方案。

它编译的证明可以在这里找到

于 2013-03-11T10:59:58.827 回答