CRTP变量inizializatin中的衍生类
考虑以下实现CRTP的代码，
#include <iostream>
template <typename T>
class base {
int n;
base() {}
friend T;
public:
explicit base(int _n) : n{_n} {}
void print() const { static_cast<const T&>(*this).whoami(); }
};
class derived : public base<derived> {
public:
void whoami() const { std::cout << n << std::endl; }
};
int main()
{
base<derived> a{6};
a.print();
}
如预期，代码打印6。
默认的构造函数base :: base（）将离开成员字段n非专业化。但是，由于它是private和base是friend derived，因此只有derived使用base :: base（）（具体来说，在其编译器生成的默认文件中）。但是派生没有公共构造函数，因此永远不会实例化。因此，从n的初始化点起，代码似乎是安全的，尽管代码的一部分实际上没有执行正确的初始化。
所以我想知道：上面的代码中是否有陷阱？
注意
请注意，它不仅仅是学术的问题。在现实情况下，base的成员可能没有明智的默认初始化，并且由base的构造函数初始化，该构建器采用一系列参数。这些参数只能在运行时确定，并在main（）中用于实例base＆lt; derived＆gt;。
class base {
public:
base(int param1, char *param2);
};
如果base没有默认构造函数，则需要一个派生的构造函数，该构建器调用base类base＆lt; derived＆gt;的构造函数。如果没有明智的默认参数，则将其称为 “随机”参数
class derived : public base<derived> {
derived() : base<derived>(42, "unexisting_file") { }
};
Consider the following code implementing CRTP
#include <iostream>
template <typename T>
class base {
int n;
base() {}
friend T;
public:
explicit base(int _n) : n{_n} {}
void print() const { static_cast<const T&>(*this).whoami(); }
};
class derived : public base<derived> {
public:
void whoami() const { std::cout << n << std::endl; }
};
int main()
{
base<derived> a{6};
a.print();
}
The code prints 6 as expected.
The default constructor base::base() would leave the member field n uninitialized. However, since it is private and base is friend to derived, only derived uses base::base() (specifically, in its compiler-generated default contructor). But derived has not public constructors, so it is never instantiated. So the code seems to be safe from the point of the initialization of n, though there is a part of the code that, effectively, does not perform the correct initialization.
So I wonder: is there a pitfall in a code like above?
Note
Please notice that the question it not merely academic. In a real-world case, the members of base may not have a sensible default initialization, and are initialized by a constructor of base that takes a series of parameters. These parameters can be possibly determined at runtime only, and used in main() to instantiate base<derived>.
class base {
public:
base(int param1, char *param2);
};
If there is no default constructor for base, you need a constructor for derived that calls the constructor of the base class base<derived>. When this has no sensible default parameters, it would be very awkward to call it with
"random" parameters
class derived : public base<derived> {
derived() : base<derived>(42, "unexisting_file") { }
};
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