我有一个指向基类foo的指针向量,它有几个子类,我想做的是基于它是哪个子类,创建一个相同实例的新类。
我之前通过一个巨大的 for 循环解决了这个问题,它使用 typeid 来找出它是什么类,但是没有办法以更一般的方式解决它吗?
基本上,我正在寻找这样的东西:
std::vector<foo*> a;
std::vector<foo*> b;
//Store a couple of classes in a and b
b[0] = new typeid(a[0]).name();
一种方法是使用一个虚拟clone()
方法,该方法返回一个指向正确类型对象的指针:
struct base
{
virtual base* clone()=0;
virtual ~base() {}
};
struct foo : public base
{
virtual base* clone() { return new foo(*this); }
};
struct bar : public base
{
virtual base* clone() { return new bar(*this); }
};
那么你的代码将是
b[0] = a[0].clone();
在现实生活中的代码中,您将返回一个智能指针,例如std::unique_ptr<base>
.
每当你有一个基于类型的巨大开关时,你可能应该使用虚函数来代替。
您应该引入某种虚clone()
函数:
#include <iostream>
#include <memory>
struct base
{
virtual ~base() {};
virtual void do_stuff() = 0;
// cloning interface
std::unique_ptr<base> clone() const
{
return std::unique_ptr<base>(do_clone());
}
private:
// actual clone implementation; uses a raw pointer to support covariance
virtual base* do_clone() const = 0;
};
struct derived_A : base
{
void do_stuff() override { std::cout << "stuff in derived_A" << std::endl; }
// purposefully hide the base implementation,
// since we know we'll be returning a derived_A
std::unique_ptr<derived_A> clone() const
{
return std::unique_ptr<derived_A>(do_clone());
}
private:
derived_A* do_clone() const override
{
return new derived_A(*this);
}
};
struct derived_B : base
{
void do_stuff() override { std::cout << "stuff in derived_B" << std::endl; }
// purposefully hide the base implementation,
// since we know we'll be returning a derived_B
std::unique_ptr<derived_B> clone() const
{
return std::unique_ptr<derived_B>(do_clone());
}
private:
derived_B* do_clone() const override
{
return new derived_B(*this);
}
};
#include <vector>
int main()
{
std::vector<std::unique_ptr<base>> v1;
std::vector<std::unique_ptr<base>> v2;
std::unique_ptr<base> x(new derived_A);
v1.push_back(std::move(x));
std::unique_ptr<base> y(new derived_B);
v1.push_back(std::move(y));
v1[0]->do_stuff();
v1[1]->do_stuff();
// clone
v2.push_back(v1[0]->clone());
v2.push_back(v1[1]->clone());
v2[0]->do_stuff();
v2[1]->do_stuff();
}
我们希望返回类型的协变(如果您持有指向静态类型的指针derived_A
,则克隆它应该产生 aderived_A
以避免冗余强制转换),这就是克隆接口被分成两部分的原因。如果std::unique_ptr<base>
与 协变,则可以一次性完成std::unique_ptr<derived>
,但这仅适用于原始指针。
我确信有一种方法可以隐藏重复的样板文件,这对读者来说是一个练习。
编辑:实际上,给你;不是太难:
#include <memory>
// Note: leaves with a public: access specifier
#define DEFINE_ABSTRACT_CLONEABLE(selfType) \
DEFINE_CLONEABLE_DETAIL(selfType) \
private: \
virtual selfType* do_clone() const = 0; \
\
public:
// Note: leaves with a public: access specifier
#define DEFINE_CLONEABLE(selfType) \
DEFINE_CLONEABLE_DETAIL(selfType) \
private: \
selfType* do_clone() const override \
{ \
return new selfType(*this); \
} \
\
public:
#define DEFINE_CLONEABLE_DETAIL(selfType) \
public: \
std::unique_ptr<selfType> clone() const \
{ \
static_assert(std::is_same<selfType, \
std::decay<decltype(*this)>::type \
>::value, \
"Must specify current class name."); \
\
return std::unique_ptr<selfType>(do_clone()); \
} \
和测试(注意较小的尺寸):
#include <iostream>
#include "cloneable.hpp" // or whatever
struct base
{
// readable error: DEFINE_ABSTRACT_CLONEABLE(int);
DEFINE_ABSTRACT_CLONEABLE(base);
virtual ~base() {};
virtual void do_stuff() = 0;
};
struct derived_A : base
{
DEFINE_CLONEABLE(derived_A);
void do_stuff() override { std::cout << "stuff in derived_A" << std::endl; }
};
struct derived_B : base
{
// error: DEFINE_CLONEABLE(derived_B);
DEFINE_ABSTRACT_CLONEABLE(derived_B);
void do_stuff() override { std::cout << "stuff in derived_B" << std::endl; }
virtual void do_thing() = 0; // abstract again
};
struct derived_AA : derived_A
{
DEFINE_CLONEABLE(derived_AA);
void do_stuff() override { std::cout << "stuff in derived_AA" << std::endl; }
};
struct derived_BB : derived_B
{
DEFINE_CLONEABLE(derived_BB);
void do_stuff() override { std::cout << "doing stuff in derived_BB" << std::endl; }
void do_thing() override { std::cout << "doing thing" << std::endl; }
};
int main()
{
std::unique_ptr<derived_AA> x(new derived_AA());
x->do_stuff();
auto xx = x->clone();
xx->do_stuff();
std::unique_ptr<derived_A> xxx = xx->clone();
xxx->do_stuff();
std::unique_ptr<base> xxxx = xxx->clone();
xxxx->do_stuff();
xxxx->clone()->do_stuff();
std::unique_ptr<derived_BB> y(new derived_BB());
y->do_stuff();
y->do_thing();
auto yy = y->clone();
yy->do_stuff();
yy->do_thing();
std::unique_ptr<derived_B> yyy = yy->clone();
yyy->do_stuff();
yyy->do_thing();
std::unique_ptr<base> yyyy = yyy->clone();
yyyy->do_stuff();
// error, lost derived information: yyyy->do_thing();
yyyy->clone()->do_stuff();
}
另一个改进是使每个新的do_clone
纯虚拟声明都强制进一步派生类来实现它,但这留给读者。
更好的方法是使用dynamic_cast
. 您可以尝试dynamic_cast
ing 然后调用适当的复制构造函数。话虽如此,我会选择clone
解决方案或尝试重新设计解决问题的方法。
例如,
child *c = dynamic_cast<child*>(base);
if(c != NULL) {
return new child(c);
}