我使用访问者模式来处理很多不同的AST问题,结果证明它工作得很好。例如,我正在使用它来检查静态类型。这在查找确切类型时效果很好,但它不适用于派生类。即如果我们有从Base
继承的Derived
,询问Derived
对象是否为Base
失败。使用访客模式检查派生类的类型?
考虑下面的C++代码:
#include <iostream>
#include <functional>
#include <memory>
using namespace std;
class Base;
class Derived;
class Visitor {
public:
virtual void visit(Base& object) = 0;
virtual void visit(Derived& object) = 0;
};
class EmptyVisitor : public Visitor {
public:
virtual void visit(Base& object) override {}
virtual void visit(Derived& object) override {}
};
template <class TYPE> class LogicVisitor : public EmptyVisitor {
public:
LogicVisitor(function<void(TYPE&)> logic) : EmptyVisitor(), logic(logic) {}
virtual void visit(TYPE& object) override { logic(object); }
private:
function<void(TYPE&)> logic;
};
class Base {
public:
virtual void accept(Visitor* visitor) {
visitor->visit(*this);
}
};
class Derived : public Base {
public:
virtual void accept(Visitor* visitor) override {
visitor->visit(*this);
}
};
template <class TYPE> bool is_type(shared_ptr<Base> base)
{
bool is_type = false;
LogicVisitor<TYPE> logic_visitor([&](TYPE& object) {
is_type = true;
});
base->accept((Visitor*)&logic_visitor);
return is_type;
}
int main() {
auto base = make_shared<Base>();
auto derived = make_shared<Derived>();
cout << "is_type<Base>(base) = " << (is_type<Base>(base) ? "true" : "false") << endl;
cout << "is_type<Derived>(base) = " << (is_type<Derived>(base) ? "true" : "false") << endl;
cout << "is_type<Base>(derived) = " << (is_type<Base>(derived) ? "true" : "false") << endl;
cout << "is_type<Derived>(derived) = " << (is_type<Derived>(derived) ? "true" : "false") << endl;
return 0;
}
它输出如预期以下结果:
is_type<Base>(base) = true
is_type<Derived>(base) = false
is_type<Base>(derived) = false
is_type<Derived>(derived) = true
虽然这是很大的检索静态类型的对象的,如何能这如果我想要is_type<Base>(derived)
返回true
而不是false
,以便我可以有效地检查类继承吗?这在C++中可能吗?
为什么你需要所有这些虚拟网,如果你可以简单地用'的std :: is_base_of'? – SergeyA
嗯,我不知道'std :: is_base_of',看起来非常有用,但问题是,这个函数是否需要RTTI? – Deathicon
@Deathicon号这是一个类型特征,在编译时进行评估。 – Rakete1111