计算函数指针的类型 [英] Computing the type of a function pointer
问题描述
请考虑以下内容:
template<typename T>
struct S
{
typedef M< &T::foo > MT;
}
这将适用于:
S<Widget> SW;
其中 Widget :: foo()
是一些功能
我将如何修改 struct S
的定义以允许以下内容:
How would I modify the definition of struct S
to allow the following instead:
S<Widget*> SWP;
推荐答案
您需要的是以下类型转换。
What you need is the following type transformation.
- 给出
T
,返回T
- 给出
T *
,返回T
- given
T
, returnT
- given
T *
, returnT
碰巧标准库已经在 std :: remove_pointer
(尽管做起来并不难)。
It so happens that the standard library already has implemented this for us in std::remove_pointer
(though it's not hard to do yourself).
这样,您就可以使用object_type = std :: remove_pointer_t< T>来写
With this, you can then write
using object_type = std::remove_pointer_t<T>;
using return_type = /* whatever foo returns */;
using MT = M<object_type, return_type, &object_type::foo>;
关于您还想使用智能指针的评论,我们必须重新定义类型
Regarding your comment that you also want to work with smart pointers, we have to re-define the type transformation.
- 给定智能指针类型
smart_ptr< T>
,返回smart_ptr< T> :: element_type
,应为T
- 给出一个指针键入
T *
,返回T
- 否则,给出
T
,返回T
本身
- given a smart pointer type
smart_ptr<T>
, returnsmart_ptr<T>::element_type
, which should beT
- given a pointer type
T *
, returnT
- otherwise, given
T
, returnT
itself
为此,我们必须编写自己的元函数。至少,我不知道标准库中有什么对这里有帮助的。
For this, we'll have to code our own meta-function. At least, I'm not aware of anything in the standard library that would help here.
我们首先定义主要的模板$
We start by defining the primary template
(the "otherwise" case).
template <typename T, typename = void>
struct unwrap_obect_type { using type = T; };
第二个(匿名)类型参数,默认为 void
将在以后使用。
The second (anonymous) type parameter that is defaulted to void
will be of use later.
对于(原始)指针,我们提供了以下部分专业化。
For (raw) pointers, we provide the following partial specialization.
template <typename T>
struct unwrap_obect_type<T *, void> { using type = T; };
如果我们在这里停止,基本上可以得到 std :: remove_pointer
。但是,我们将为智能指针添加额外的部分专业化。当然,我们首先必须定义什么是智能指针。就本示例而言,我们将使用嵌套的 typedef
名为 element_type
的每个类型作为智能指针。
If we'd stop here, we'd basically get std::remove_pointer
. But we'll add an additional partial specialization for smart pointers. Of course, we'll first have to define what a "smart pointer" is. For the purpose of this example, we'll treat every type with a nested typedef
named element_type
as a smart pointer. Adjust this definition as you see fit.
template <typename T>
struct unwrap_obect_type
<
T,
std::conditional_t<false, typename T::element_type, void>
>
{
using type = typename T::element_type;
};
第二种类型参数 std :: conditional_t< false,类型名称T :: element_type,void>
是一种模拟 <$ c $的复杂方法c> 14中的c> std :: void_t 。这个想法是,我们具有以下部分类型函数。
The second type parameter std::conditional_t<false, typename T::element_type, void>
is a convoluted way to simulate std::void_t
in C++14. The idea is that we have the following partial type function.
- 给出了类型
T
嵌套名为element_type
的嵌套typedef
,返回void
- 否则,触发替换失败
- given a type
T
with a nestedtypedef
namedelement_type
, returnvoid
- otherwise, trigger a substitution failure
因此,如果我们要处理智能指针,我们将获得比主要模板
更好的匹配,否则,SFINAE将从进一步考虑中删除此部分专业化。
Therefore, if we are dealing with a smart pointer, we'll get a better match than the primary template
and otherwise, SFINAE will remove this partial specialization from further consideration.
这是一个可行的例子。 TC 建议使用 std :: mem_fn
来调用成员函数。
Here is a working example. T.C. has suggested using std::mem_fn
to invoke the member function. This makes the code a lot cleaner than my initial example.
#include <cstddef>
#include <functional>
#include <iostream>
#include <memory>
#include <string>
#include <utility>
template <typename ObjT, typename RetT, RetT (ObjT::*Pmf)() const noexcept>
struct M
{
template <typename ThingT>
static RetT
call(ThingT&& thing) noexcept
{
auto wrapper = std::mem_fn(Pmf);
return wrapper(std::forward<ThingT>(thing));
}
};
template <typename T, typename = void>
struct unwrap_obect_type { using type = T; };
template <typename T>
struct unwrap_obect_type<T *, void> { using type = T; };
template <typename T>
struct unwrap_obect_type<T, std::conditional_t<false, typename T::element_type, void>> { using type = typename T::element_type; };
template <typename T>
struct S
{
template <typename ThingT>
void
operator()(ThingT&& thing) const noexcept
{
using object_type = typename unwrap_obect_type<T>::type;
using id_caller_type = M<object_type, int, &object_type::id>;
using name_caller_type = M<object_type, const std::string&, &object_type::name>;
using name_length_caller_type = M<object_type, std::size_t, &object_type::name_length>;
std::cout << "id: " << id_caller_type::call(thing) << "\n";
std::cout << "name: " << name_caller_type::call(thing) << "\n";
std::cout << "name_length: " << name_length_caller_type::call(thing) << "\n";
}
};
class employee final
{
private:
int id_ {};
std::string name_ {};
public:
employee(int id, std::string name) : id_ {id}, name_ {std::move(name)}
{
}
int id() const noexcept { return this->id_; }
const std::string& name() const noexcept { return this->name_; }
std::size_t name_length() const noexcept { return this->name_.length(); }
};
int
main()
{
const auto bob = std::make_shared<employee>(100, "Smart Bob");
const auto s_object = S<employee> {};
const auto s_pointer = S<employee *> {};
const auto s_smart_pointer = S<std::shared_ptr<employee>> {};
s_object(*bob);
std::cout << "\n";
s_pointer(bob.get());
std::cout << "\n";
s_smart_pointer(bob);
}
这篇关于计算函数指针的类型的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!