派生类构造函数中的数据驱动标志设置 [英] Data-driven flag setting in derived class constructor

问题描述

说我有一个基类，里面有一个标志，派生类必须设置：

  struct Base 
 {
 bool flag; 
 Base（bool flag）：flag（flag）{} 
}; 
  

我想配置哪些派生类将标志设置为 true / false 以数据驱动的方式 - 即我想从标题中配置。

  struct Derived1：Base 
 {
 Derived1（）：Base（expr）{} 
}; 
  

其中 expr 是 / em>（不知道什么）能够从标题获取信息 - 告诉是否 Derived1 应该使标志 true或false。理想情况下，如果我创建一个新的派生类，但是没有指定标题中的标志，我会收到错误，但这并不是强制性的。这样我可以修改一个单一的中央位置来进行更改。

这是什么习惯的方法？

解决方案

使用单个功能的替代版本可能更紧凑：

  struct Derived1 ：Base 
 {
 Derived1（）：Base（theFlag（this））{} 
}; 
  

然后在标题中：

 模板< typename T> 
 bool theFlag（T *）
 {
 if（typeid（T）== typeid（Derived1））return true; 
 if（typeid（T）== typeid（Derived2））return false; 
 if（typeid（T）== typeid（Derived3））return true; 
 
 throw std :: runtime_error（No theFlag is given for this type）; 
} 
  

如果你结婚到编译时检查，最好的办法要引入一些重复：

 模板< typename T> 
 bool theFlag（T *）
 {
 static_assert（
 std :: is_same< T，Derived1> :: value || 
 std :: is_same< T， Derived2> :: value || 
 std :: is_same< T，Derived3> :: value，
没有给这个类型的
; 
 
 if（typeid（T）== typeid（Derived1））return true; 
 if（typeid（T）== typeid（Derived2））return false; 
 if（typeid（T）== typeid（Derived3））return true; 
} 
  

这基本上依赖于SFINAE - 编译器将无法找到重载对于 theFlag ，如果您使用不支持的参数调用它，基本上是。

Say I have a base class with a flag inside of it which derived classes have to set:

struct Base
{
    bool flag;
    Base(bool flag):flag(flag) {}
};

I want to configure which derived classes set the flag to true/false in a data-driven way - i.e. I'd like to configure this from a header.

struct Derived1 : Base
{
    Derived1() : Base( expr ) {}
};

Where expr is something (don't know what yet) that is able to get the info from the header - tell whether Derived1 should make flag true or false. Ideally, I'd get an error if I make a new derived class but fail to specify the flag in the header, but this isn't mandatory. This way I can just modify a single central location to make changes.

What's the idiomatic approach for this?

解决方案

An alternative version that uses a single function might be more compact:

struct Derived1 : Base
{
    Derived1() : Base(theFlag(this)) {}
};

Then in the header:

template <typename T>
bool theFlag(T*)
{
   if (typeid(T) == typeid(Derived1)) return true;
   if (typeid(T) == typeid(Derived2)) return false;
   if (typeid(T) == typeid(Derived3)) return true;

   throw std::runtime_error("No theFlag is given for this type");
}

If you are married to the compile-time check, the best you could do is to introduce a bit of duplication:

template <typename T>
bool theFlag(T*)
{
   static_assert(
      std::is_same<T, Derived1>::value ||
      std::is_same<T, Derived2>::value ||
      std::is_same<T, Derived3>::value,
      "No theFlag is given for this type"
   );

   if (typeid(T) == typeid(Derived1)) return true;
   if (typeid(T) == typeid(Derived2)) return false;
   if (typeid(T) == typeid(Derived3)) return true;
}

This basically relies on SFINAE - the compiler would not be able to find an overload for theFlag if you called it with an unsupported argument, essentially.

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