問題描述
我在 C++ 中有這樣一個模板
I have such a template in C++
template<typename T, T* P> struct Ptr {};
所以我可以這樣使用它:
so I can use it as such:
const int i = 0;
Ptr<int, &i> ptr;
或
Ptr<decltype(i), &i> ptr;
但我不想指定類型 int 或身份 i 兩次，我只想使用
But I don't want to specify the type int or identity i twice, I want to use just
Ptr<&i> ptr;
并讓編譯器自己找出 int 類型部分.
and let the compiler figure out the int type part by itself.
我如何聲明我的模板來做到這一點?
How can I declare my template to do that ?
我讀過這個問題，但答案是使用宏，這不好:模板c++的模板?
I've read this question but the answer is using macros, that's not nice: template of template c++?
我可以通過沒有宏的模板來做到這一點嗎?我使用的是 Visual C++ 2013.
can I do this by just template without macros ? I'm using Visual C++ 2013.
推薦答案
UPDATE
c++17 引入了P0127R2 使用 auto 聲明非類型模板參數"，允許聲明非類型模板使用 auto 作為實際類型占位符的參數:
c++17 introduced "P0127R2 Declaring non-type template parameters with auto", allowing to declare a non-type template parameter(s) with auto as a placeholder for the actual type:
template <auto P> struct Ptr {};
也就是說，P 是一個非類型模板參數.它的類型可以通過 decltype(P) 推斷出來.
That is, P is a non-type template parameter. Its type can be inferred with decltype(P).
auto 遵循眾所周知的推導和偏序規則.在您的情況下，可以將類型限制為僅接受指針:
auto in a template parameter list is subject to well-known deduction and partial ordering rules. In your case, the type can be constrained to accept pointers only:
template <auto* P> struct Ptr {};
請注意，即使對于更詳細的檢查，使用 auto 的語法也足夠了，例如:
Note that the syntax utilizing auto is sufficient even for more detailed inspection, e.g.:
template <typename F>
struct FunctionBase;
template <typename R, typename... Args>
struct FunctionBase<R(*)(Args...)> {};
template <auto F>
struct Function : FunctionBase<decltype(F)> {};
也可以使用推斷類型作為其他模板參數的約束:
It's also possible to use the inferred type as a contraint for other template parameters:
template <auto I, decltype(I)... Is>
struct List {};
<小時>
舊答案
既然您問的是一個沒有宏定義幫助的基于類模板的純解決方案，那么答案很簡單:至于現在(2014 年 12 月，c++14) 不可能.
Since you are asking about a pure class template-based solution without the help of macro definitions then the answer is simple: as for now (Dec 2014, c++14) it is not possible.
這個問題已經被 WG21 C++ 標準委員會確定為需要，并且有幾個建議讓模板自動推斷非類型模板參數的類型.
This issue has been already identified by the WG21 C++ Standard Committee as a need and there are several proposals to let templates automatically infer the type of non-type template arguments.
最接近的是N3601 隱式模板參數:
此示例的目的是消除對冗余 template 習語的需要.這個習語被廣泛使用，在谷歌上的點擊量超過 10 萬次.
Implicit template parameters
The purpose of this example is to eliminate the need for the redundant
template<typename T, T t>idiom. This idiom is widely used, with over 100k hits on Google.
目標是能夠替換像template 和另一個聲明，這樣我們就可以像 C<&X::f> 一樣實例化模板，而不必說 C.
The goal is to be able to replace a template declaration like template<typename T, T t> struct C; with another declaration so that we can instantatiate the template like C<&X::f> instead of having to say C<decltype(&X::f), &X::f>.
基本思想是能夠說template