如何提取可变参数函数的一组选定参数并使用它们调用另一个函数 [英] How to extract a selected set of arguments of a variadic function and use them to call another function
问题描述
我有一个可变参数函数 zoo ，该函数带有N个参数，其中N在编译时是已知的（它是包含该函数的类的模板参数）。
I have a variadic function zoo which takes N arguments, where N is known at compile time (it is a template parameter of the class containing the function).
template <int N>
struct B
{
template <typename... Args>
static void zoo(Args... args)
{
static_assert(size of...(args) == N, "");
// do something
}
};
我还有另一个可变参数函数 foo ，该函数接受M个参数，其中M> N，并且在编译时已知（它是包含函数的类的模板参数）。我有一个静态index_array，其中包含要传递给 zoo 的 foo 的参数的索引。
I have another variadic function foo which takes M arguments, where M>N and is known at compile time (it is a template parameter of the class containing the function). I have a static index_array containing the indices of the arguments of foo I want to pass to zoo.
从 foo 的正文中，我想调用 zoo 并传递参数的选定子集foo 。
From the body of foo I want to call zoo passing a selected subset of the arguments of foo.
执行此操作的最佳方法是什么？理想情况下实现完美的内联，即使所有内容都编译成一条指令而没有任何函数指针间接？
What is the best way to do this? Ideally achieving perfect inlining, i.e. so that everything is compiled into just one instruction with no function pointers indirections?
template<int...I>
struct indices
{
static constexpr int N = sizeof...(I);
};
template <int M, typename...X>
struct A
{
// here I am simplifying, in reality IS will be built at compile time based on X
typedef indices<0,2,3> IS;
template <typename... Args>
static void foo(Args... args)
{
static_assert(size of...(args) == M, "");
// do some magic to achieve the function call described in pseudo-code
// B<IS::N>::zoo(args(IS(0),IS(1),IS(2)))
// ideally this should be perfectly inlined to just have the call above
}
};
请注意，上面的代码是我的问题的简化，旨在说明问题。
Please note the code above is a simplification of my problem, designed for the purpose of illustrating the question.
编辑：
如下所述，我描述了用例：
我正在使用基于模板的库来驱动微控制器引脚。微控制器具有多个端口（可作为内存中的字节访问），每个端口最多具有8个引脚（位）。通过模板参数X，A类是一堆针脚，其中每个针脚都定义为Pin。 B类操作同一端口上的所有引脚。 A :: foo是用于修改某些引脚的函数，其参数的顺序与X模板参数包中指定引脚的顺序相同。 foo需要按端口对参数进行分组，然后分派到代表各个端口的B类中，在这些类中，所有参数都在一条指令中融合并写入控制器。
As asked below, I describe the use case: I am playing with a template based library to drive micro-controller pins. A micro controller has several ports (accessible as bytes in memory) and each port has up to 8 pins (bits). Class A is a bundle of pins via the template argument X, where every pin is defined as Pin. Class B manipulates all pins on the same port. A::foo is a function to modify some of the pins, with arguments in the same order as the order with which the pins are specified in the X template argument pack. foo needs to group the arguments by ports and dispatch to the B classes which representing individual ports, where all arguments are fused and written to the controller in a single instruction.
推荐答案
您可以创建一个助手来提取 nth_arg ，如下所示：
You can create a helper to extract the nth_arg like this:
template <int I>
struct ignore
{
template <typename T>
ignore(T&&) // This constructor accepts anything
{
}
};
template <typename T>
struct nth_arg;
template <size_t... DropIndexes>
struct nth_arg<std::integer_sequence<size_t, DropIndexes...>>
{
template <typename Arg, typename... Rest>
static decltype(auto) get(ignore<DropIndexes>..., // ignore args 0...n-1
Arg&& arg,
Rest&&...) // also ignore the rest
{
return std::forward<Arg>(arg); // return nth arg
}
};
然后致电
template <int... Is, typename... Args>
static void call_zoo(indices<Is...>, Args&&... args)
{
B<sizeof...(Is)>::zoo(nth_arg<std::make_index_sequence<Is>>::get(
std::forward<Args>(args)...)...);
}
template <int M>
struct A
{
typedef indices<0, 2, 3> IS;
template <typename... Args>
static void foo(Args... args)
{
static_assert(sizeof...(args) == M, "");
call_zoo(IS{}, std::forward<Args>(args)...);
}
};
如果您使用的是C ++ 11，则可以轻松滚动自己的 integer_sequence 。
If you're using C++11, you can easily roll your own integer_sequence.
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