4

是否有任何性能或稳健性的理由偏爱其中一种?

#include <iostream>
#include <typeinfo>

struct B
{
    virtual bool IsType(B const * b) const { return IsType2nd(b) && b->IsType2nd(this); }
    virtual bool IsType2nd(B const * b) const { return dynamic_cast<decltype(this)>(b) != nullptr; }
};

struct D0 : B
{
    virtual bool IsType(B const * b) const { return IsType2nd(b) && b->IsType2nd(this); }
    virtual bool IsType2nd(B const * b) const { return dynamic_cast<decltype(this)>(b) != nullptr; }
};

struct D1 : B
{
    virtual bool IsType(B const * b) const { return IsType2nd(b) && b->IsType2nd(this); }
    virtual bool IsType2nd(B const * b) const { return dynamic_cast<decltype(this)>(b) != nullptr; }
};

int main()
{
    using namespace std;
    B b, bb;
    D0 d0, dd0;
    D1 d1, dd1;

    cout << "type B  == type B  : " << (b.IsType(&bb)   ? "true " : "false") << endl;
    cout << "type B  == type D0 : " << (b.IsType(&dd0)  ? "true " : "false") << endl;
    cout << "type B  == type D1 : " << (b.IsType(&dd1)  ? "true " : "false") << endl;
    cout << "type D0 == type B  : " << (d0.IsType(&bb)  ? "true " : "false") << endl;
    cout << "type D0 == type D0 : " << (d0.IsType(&dd0) ? "true " : "false") << endl;
    cout << "type D0 == type D1 : " << (d0.IsType(&dd1) ? "true " : "false") << endl;
    cout << "type D1 == type B  : " << (d1.IsType(&bb)  ? "true " : "false") << endl;
    cout << "type D1 == type D0 : " << (d1.IsType(&dd0) ? "true " : "false") << endl;
    cout << "type D1 == type D1 : " << (d1.IsType(&dd1) ? "true " : "false") << endl;
    cout << endl;
    cout << "type B  == type B  : " << (typeid(b) == typeid(bb)   ? "true " : "false") << endl;
    cout << "type B  == type D0 : " << (typeid(b) == typeid(dd0)  ? "true " : "false") << endl;
    cout << "type B  == type D1 : " << (typeid(b) == typeid(dd1)  ? "true " : "false") << endl;
    cout << "type D0 == type B  : " << (typeid(d0) == typeid(&bb) ? "true " : "false") << endl;
    cout << "type D0 == type D0 : " << (typeid(d0) == typeid(dd0) ? "true " : "false") << endl;
    cout << "type D0 == type D1 : " << (typeid(d0) == typeid(dd1) ? "true " : "false") << endl;
    cout << "type D1 == type B  : " << (typeid(d1) == typeid(bb)  ? "true " : "false") << endl;
    cout << "type D1 == type D0 : " << (typeid(d1) == typeid(dd0) ? "true " : "false") << endl;
    cout << "type D1 == type D1 : " << (typeid(d1) == typeid(dd1) ? "true " : "false") << endl;
}

输出:

type B  == type B  : true 
type B  == type D0 : false
type B  == type D1 : false
type D0 == type B  : false
type D0 == type D0 : true 
type D0 == type D1 : false
type D1 == type B  : false
type D1 == type D0 : false
type D1 == type D1 : true 

type B  == type B  : true 
type B  == type D0 : false
type B  == type D1 : false
type D0 == type B  : false
type D0 == type D0 : true 
type D0 == type D1 : false
type D1 == type B  : false
type D1 == type D0 : false
type D1 == type D1 : true 
4

2 回答 2

1

从设计的角度来看,双重调度要灵活得多:

  • 目前,您检查类型与IsType2nd(b) && b->IsType2nd(this). 但可能在某些时候你想进一步推导

  • 但是有一天您可能想要进一步派生 D1,但在比较类型时仍想将其视为 D1 对象的位置。这种特殊情况很容易通过双重调度完成。

这种灵活性是有代价的:汇编代码将通过 vtable 使用 2 个间接调用,外加一个动态转换。

直接类型信息并不是最好的设计,正如 Sergey 指出的那样:它始终是严格的类型比较,不可能有特殊情况。

这种不灵活性带来了代码生成简单的优势:代码只需在 vtable 的开头获取动态类型信息(并且编译器可以轻松地针对编译时已知类型的对象优化此提取。

出于好奇,这里生成了一些代码:他的 typeid 在编译时被优化掉了,而双重调度仍然依赖于间接调用。

于 2016-02-12T21:40:02.690 回答
0

如评论中所述,它遵循另一种既不使用typeid也不依赖于dynamic_cast.

我添加了几个额外的示例来展示如何轻松定义类型(例如,这里两者D1和都D1Bis显示为相同的族类型,即使它们实际上是不同的类型)。
无论如何,不​​确定这是一个理想的功能......

希望它对你感兴趣。

#include<iostream>

struct BB {
    virtual unsigned int GetType() = 0;

    bool IsType(BB *other) {
        return GetType() == other->GetType();
    }

protected:
    static unsigned int bbType;
};

unsigned int BB::bbType = 0;

struct B: public BB {
    unsigned int GetType() override {
        static unsigned int bType = BB::bbType++;
        return bType;
    }
};

struct D0: public B {
    unsigned int GetType() override {
        static unsigned int d0Type = BB::bbType++;
        return d0Type;
    }
};

struct D1: public B {
    unsigned int GetType() override {
        static unsigned int d1Type = BB::bbType++;
        return d1Type;
    }
};

struct D1Bis: public D1 { };

int main() {
    using namespace std;
    B b, bb;
    D0 d0, dd0;
    D1 d1, dd1;
    D1Bis d1Bis;

    cout << "type B  == type B  : " << (b.IsType(&bb)   ? "true " : "false") << endl;
    cout << "type B  == type D0 : " << (b.IsType(&dd0)  ? "true " : "false") << endl;
    cout << "type B  == type D1 : " << (b.IsType(&dd1)  ? "true " : "false") << endl;
    cout << "type B  == type D1BIS : " << (b.IsType(&d1Bis)  ? "true " : "false") << endl;
    cout << "type D0 == type B  : " << (d0.IsType(&bb)  ? "true " : "false") << endl;
    cout << "type D0 == type D0 : " << (d0.IsType(&dd0) ? "true " : "false") << endl;
    cout << "type D0 == type D1 : " << (d0.IsType(&dd1) ? "true " : "false") << endl;
    cout << "type D0 == type D1BIS : " << (d0.IsType(&d1Bis) ? "true " : "false") << endl;
    cout << "type D1 == type B  : " << (d1.IsType(&bb)  ? "true " : "false") << endl;
    cout << "type D1 == type D0 : " << (d1.IsType(&dd0) ? "true " : "false") << endl;
    cout << "type D1 == type D1 : " << (d1.IsType(&dd1) ? "true " : "false") << endl;
    cout << "type D1 == type D1Bis : " << (d1.IsType(&d1Bis) ? "true " : "false") << endl;
}
于 2016-02-14T00:05:55.713 回答