SFINAE尝试使用bool给出编译器错误：“模板参数'T :: value'涉及模板参数” [英] SFINAE tried with bool gives compiler error: "template argument ‘T::value’ involves template parameter"

问题描述

我尝试使用 bool 实现SFINAE（不像流​​行的 void _ trick ）：

I tried to implement an SFINAE using bool (unlike popular void_ trick):

  template<typename T, bool = true>
  struct Resolve
  {
    static const bool value = false;
  };

  template<typename T>
  struct Resolve<T, T::my_value>
  {
    static const bool value = true;
  };

目标是专门化，类 static const bool my_value = true; 里面定义。如果它们被定义 false 或没有定义，那么不专门化它。也就是

The goal is to specialize, the classes which have static const bool my_value = true; defined inside it. If they are defined false or not defined then don't specialize it. i.e.

struct B1 {  // specialize Resolve for this case
  static const bool my_value = true;
};
struct B2 {  // don't specialize
  static const bool my_value = false;
};
struct B3 {};  // don't specialize

当在 B1 它给出编译错误：

When applying the above trick on B1 it gives the compilation error:

Resolve<B1>::value;

错误：模板参数'T :: my_value'涉及模板参数s）

error: template argument ‘T::my_value’ involves template parameter(s)

我知道这可以用替代方法来实现。但是，我有兴趣知道，为什么它在这里给出编译器错误，可以在这个代码本身解决？

I am aware that this can be achieved with alternate ways. However, I am interested in knowing, why it gives compiler error here and can it be solved in this code itself ?

推荐答案

你正在做的是禁止的部分§14.5.4/ 9说，

Actually what you're doing is forbidden by section §14.5.4/9 which says,

部分专用非类型参数除非参数表达式是简单标识符，否则表达式不应涉及部分专门化的模板参数。

诀窍可能是使用类型用于第二个模板参数， -type 值，如下所示：

The trick could be using a type for second template parameter as well, encapsulating the non-type value, as described below:

template<bool b> struct booltype {};

template<typename T, typename B = booltype<true> >
struct Resolve
{
  static const bool value = false;
};

template<typename T>
struct Resolve<T, booltype<T::my_value> >
{
  static const bool value = true;
};

现在编译罚款。

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