是否可以模拟模板< auto X>？ [英] Is it possible to emulate template<auto X>?

问题描述

它有可能吗？我想要启用编译时传递参数。假设它仅仅为了用户方便，因为人们总是可以用 template< class T，T X> 来输出真实类型，但是对于一些类型， - 函数，这是非常乏味，即使以 decltype 作为快捷方式。考虑下面的代码：

Is it somehow possible? I want that to enable compile-time passing of arguments. Suppose it's only for user convenience, as one could always type out the real type with template<class T, T X>, but for some types, i.e. pointer-to-member-functions, it's pretty tedious, even with decltype as a shortcut. Consider the following code:

struct Foo{
  template<class T, T X>
  void bar(){
    // do something with X, compile-time passed
  }
};

struct Baz{
  void bang(){
  }
};

int main(){
  Foo f;
  f.bar<int,5>();
  f.bar<decltype(&Baz::bang),&Baz::bang>();
}

是否可以将其转换为以下内容？

Would it be somehow possible to convert it to the following?

struct Foo{
  template<auto X>
  void bar(){
    // do something with X, compile-time passed
  }
};

struct Baz{
  void bang(){
  }
};

int main(){
  Foo f;
  f.bar<5>();
  f.bar<&Baz::bang>();
}

推荐答案

在C ++中没有这样的功能。最接近的是宏：

After your update: no. There is no such functionality in C++. The closest is macros:

#define AUTO_ARG(x) decltype(x), x

f.bar<AUTO_ARG(5)>();
f.bar<AUTO_ARG(&Baz::bang)>();

---

听起来像你想要一个发电机：

---

Sounds like you want a generator:

template <typename T>
struct foo
{
    foo(const T&) {} // do whatever
};

template <typename T>
foo<T> make_foo(const T& x)
{
    return foo<T>(x);
}

现在代替拼写：

foo<int>(5);

您可以：

make_foo(5);

推导参数。

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