C++ static template member, one instance for each

2019-01-07 22:55发布

Usually static members/objects of one class are the same for each instance of the class having the static member/object. Anyways what about if the static object is part of a template class and also depends on the template argument? For example, like this:

template<class T>
class A{
public:
  static myObject<T> obj;
}

If I would cast one object of A as int and another one as float, I guess there would be two obj, one for each type?

If I would create multiple objects of A as type int and multiple floats, would it still be two obj instances, since I am only using two different types?

4条回答
我欲成王,谁敢阻挡
2楼-- · 2019-01-07 23:07

There are as many static member variables as there are classes and this applies equally to templates. Each separate instantiation of a template class creates only one static member variable. The number of objects of those templated classes is irrelevant.

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神经病院院长
3楼-- · 2019-01-07 23:07

In C++ templates are actually copies of classes. I think in your example there would be one static instance per template instance.

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▲ chillily
4楼-- · 2019-01-07 23:14

A<int> and A<float> are two entirely different types, you cannot cast between them safely. Two instances of A<int> will share the same static myObject though.

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再贱就再见
5楼-- · 2019-01-07 23:24

Static members are different for each diffrent template initialization. This is because each template initialization is a different class that is generated by the compiler the first time it encounters that specific initialization of the template.

The fact that static member variables are different is shown by this code:

#include <iostream>

template <class T> class Foo {
  public:
    static int bar;
};

template <class T>
int Foo<T>::bar;

int main(int argc, char* argv[]) {
  Foo<int>::bar = 1;
  Foo<char>::bar = 2;

  std::cout << Foo<int>::bar  << "," << Foo<char>::bar;
}

Which results in

1,2
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