将 std::tuple 用于模板参数列表而不是类型列表
我正在尝试调用这样的模板函数:
I'm trying to make a call to a templated function like this :
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，因为我对这个函数进行了很多调用，而且元组很长:
I know I could "simply" pass the types of the tuple. This is what I do know but it is quite cumbersome since I make a lot of calls to this function and since the tuples are quite long:
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:
So I tried multiple implementations of the get method, but with no success :
#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.
I don't get why is there a missing template parameter.
所以我尝试了另一种实现:
So I tried another implementation :
#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
}
然后我得到这个错误:
// ----- 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.
Again, I don't understand the errors due to a missing identifier.
我现在想知道我想要实现的目标是否可能.是否可以像我想要的那样使用 std::tuple ?还是有更好的办法?
I'm wondering now if what I want to achieve is even possible. Is it possible to use a std::tuple like I want to ? Or is there a better way ?
推荐答案
您的第一个解决方案失败了，因为 get 的第二个重载在其自己的返回类型声明处不可见；要解决这个问题，您需要将返回类型计算分离到它自己的子程序中.
Your first solution is failing because the second overload to get is not visible at the point of its own return type declaration; to get around this you would need to separate out the return type computation into its own subprogram.
第二种方案更接近；问题是你只是在推断 template std::tuple，而不是它的参数.推断可变参数(例如，tuple 的类型参数)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:
The second solution is closer; the problem is that you're only inferring the template std::tuple, not its arguments. An easy way to infer variadic arguments (e.g. type arguments to tuple) is through an empty tag structure, requiring one extra level of indirection:
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:
You should check to see whether you can write the solution using pack expansion instead of recursion, for example:
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>{});
}
};
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