C ++中类模板的模板参数推导17：我做错了吗？ [英] Template argument deduction for class templates in C++17: am I doing it wrong?

问题描述

根据
https://gcc.gnu.org/projects /cxx-status.html ，与标记 -std = c ++ 1z 一起使用的g ++版本7支持类模板的模板参数推导。 p>

我期望编译下面的代码，特别是 Base 是一个抽象类，因此：

1.编译器知道可以创建 Base 的实例;

2.指向base pt_base 指向一个明确定义的实例（即 Derived {42} ），其中类型（ int

 模板< typename ValueType> 
 class Base {
 public：
 virtual ValueType getValue（）= 0; 
}; 
 
模板< typename ValueType> 
 class Derived：public Base< ValueType> {
 public：
 Derived（ValueType argt）{value = argt; } 
 virtual ValueType getValue（）{return value; } 
 ValueType值; 
}; 
 
 int main（）{
 Base * pt_base = new（Derived< int> {42}）; // *错误* 
 delete pt_base; 
} 
  

然而，不会编译。 G ++抱怨：模板占位符类型'Base'后面必须跟一个简单的声明符id ;如果我理解正确，它不会推导出模板参数。

很遗憾，因为我想动态地决定哪个派生类 pt_base 指向（可以是来自 Derived< someType> 或来自类 Derived2< someType2> ）的对象。这样，一个数组或 vector< Base *> 可以存储指向各种派生类的对象的指针。

GCC只有C ++ 17的实验支持，我不有权访问另一个编译器，所以虽然我得到一个编译错误，我不知道我的代码是错误的。你认为什么？

我们如何动态地决定 pt_base 指向一个来自 Derived< someType> 或 Derived2 （所以可以使用多态性）？

解决方案




或新表达式：

  auto p = new派生（42）; 
  

或者功能风格的转换：

  FOO（导出的（42））; 
  

它不适用于声明指针。

---

您必须简单地提供模板参数，就像您一直需要的那样。或者，我想：

  template< class T> Base< T> * downcast（Base< T> * p）{return p; } 
 auto pt_base = downcast（new Derived（42））; 
  

According to https://gcc.gnu.org/projects/cxx-status.html, version 7 of g++, used with flag -std=c++1z, supports template argument deduction for class templates.

I would expect the following code to compile, especially as Base is an abstract class, therefore:
1. the compiler knows no instance of Base can be created;
2. the pointer to base pt_base points to a clearly defined instance (i.e. Derived<int>{42}) where the type (int) is explicit.

template<typename ValueType>
class Base {
public:
    virtual ValueType getValue() = 0;
};

template<typename ValueType>
class Derived : public Base<ValueType>{
public:
    Derived(ValueType argt){ value = argt; }
    virtual ValueType getValue(){ return value; }
    ValueType value;
};

int main(){
    Base *pt_base = new(Derived<int>{42}); // *ERROR*
    delete pt_base;
}

Yet, it does not compile. G++ complains that "template placeholder type 'Base' must be followed by a simple declarator-id"; if I understand correctly, it does not deduce the template argument.
It's a pity because I would like to dynamically decide which derived class pt_base points to (could be an object from class Derived<someType> or from class Derived2<someType2>). That way, an array or a vector<Base *> could store pointers to objects of various derived classes.

GCC only has experimental support for C++17 and I don't have access to another compiler, so although I get a compile error I am not sure my code is wrong. What do you think?
And how could we dynamically decide that pt_base points to an object from either Derived<someType> or Derived2<someType2> (so polymorphism can be used)?

解决方案

Class template argument deduction works for declaring instances of class types:

Derived d(42);

Or new-expressions:

auto p = new Derived(42);

Or function-style casts:

foo(Derived(42));

It does not work for declaring pointers.

---

You'll have to simply provide the template arguments as you've always had to. Or, I guess:

template <class T> Base<T>* downcast(Base<T>* p) { return p; }
auto pt_base = downcast(new Derived(42));

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