是否存在任何性能或稳健性原因而不是优先于另一个?
#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
最佳答案 从设计角度来看,双重调度更加灵活:
>目前,您检查类型与IsType2nd(b)&&的严格因素. B-> IsType2nd(本).但可能在某个时候你想进一步衍生出来
>但是有一天你可能想要进一步推导出D1,但仍然想把它当作比较类型的D1对象.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:汇编程序代码将通过vtable使用2个间接调用,以及动态强制转换.
直接类型信息并不是最好的设计,正如谢尔盖所指出的那样:它将始终是一种严格的类型比较,没有特殊情况可能.
这种不灵活性有利于代码生成的简单性:代码只需要在vtable的开头获取动态类型信息(并且编译器可以很容易地优化这种类型在编译时已知类型的对象.
为了好奇,这里some code generated:他的typeid在编译时被优化掉了,而double displatch仍然依赖于间接调用.