CRTP：将类型从派生类传递到基类
在CRTP 中，基类可以使用派生类的函数和变量。但是，派生类的类型不能直接被基类使用，见下面的代码：
#include <iostream>
template <class Derived>
class A {
public:
//using Scalar = typename Derived::Scalar; // Error!
static constexpr int NA1 = Derived::NB1;
static constexpr int NA2 = Derived::NB2;
static constexpr int NA3 = Derived::NB3;
};
template <int _N = 2>
class B : public A<B<_N>> {
public:
using Scalar = double;
static constexpr int NB1 = 1;
static constexpr int NB2 = _N;
static constexpr int NB3 { sizeof(Scalar) };
};
int main(int argc, char** argv)
{
using Type = B<2>;
std::cout << Type::NA1 << ' '
<< Type::NA2 << ' '
<< Type::NA3 << '\n';
}
// output:
// 1 2 8
如果using Scalar = typename Derived::Scalar;取消注释该行，则会发生错误：
main.cpp:6:11: error: invalid use of incomplete type 'class B<2>'
我知道类型 ( Scalar) 可以作为模板参数传递给基类，但为什么不能像变量一样使用它呢？这只是语言规则吗？或者是否有任何逻辑限制使这无法实现？
回答
Inside A, B is an incomplete type - a compiler hasn't yet seen the complete declaration of B, so you can't use Scalar inside the declaration of A. This is a natural restriction.
The difference between a type and a scalar in your example arises because instantiation of initialization of NA happens not at the point of declaration, but only after B was seen by a compiler (and became a complete type).
Let's change the code and force a compiler to use NA value inside the class declaration:
template <class Derived>
class A {
public:
static constexpr int NA1 = Derived::NB1;
std::array<int, NA1> foo();
};
Now you'll get essentially the same error:
template <class Derived>
class A {
public:
static constexpr int NA1 = Derived::NB1;
std::array<int, NA1> foo();
};
This is similar to member functions: you can't use a CRTP base type in their declarations, but you can use that type in their bodies:
void foo() {
std::array<int, NA1> arr;
// ...
}
将编译，因为实例化发生在基类已经是完整类型的地方。