我想做的是这样的事情:
void SomeFunction(Base *){//do something}
void SomeFunction(Derived *){//do something else}
这在C ++中是否可行?到目前为止,我的尝试只是调用函数的基本版本。 这是一个清晰的例子。
示例:
#include <iostream>
#include <vector>
class Base
{
public:
Base () {std::cout << "Base Constructor for " << this << std::endl;}
void virtual PrintSomething ()
{
std::cout << "I am Base!" << std::endl;
};
};
class Derived : public Base
{
public:
Derived () : Base () {std::cout << "Derived Construtor for " << this << std::endl;}
void virtual PrintSomething ()
{
std::cout << "I am Derived!" << std::endl;
};
};
void DoAmazingStuff ( Base * ) {std::cout << "Amazing!" << std::endl;}
void DoAmazingStuff ( Derived * ) {std::cout << "DERP!" << std::endl;}
int main ()
{
std::vector<Base *> some_vector;
Base *x = new Base ();
Derived *y = new Derived ();
some_vector.push_back ( x );
some_vector.push_back ( y );
// This is the part that isn't functioning as expected.
/******************************************/
DoAmazingStuff ( some_vector[0] );
DoAmazingStuff ( some_vector[1] );
/******************************************/
std::cin.get ();
std::cin.get ();
delete some_vector[0];
delete some_vector[1];
}
永远不会调用以下行:
void DoAmazingStuff ( Derived * ) {std::cout << "DERP!" << std::endl;}
答案 0 :(得分:7)
当然有可能;但只有在指针的静态类型为Derived
时才会调用Derived*
版本。例如:
Base * b = new Base; // static and dynamic types are Base
Derived * d = new Derived; // static and dynamic types are Derived
Base * bd = new Derived; // static type Base, dynamic type Derived
SomeFunction(b); // Base overload
SomeFunction(d); // Derived overload
SomeFunction(bd); // Base overload (from static type)
动态调度(基于动态类型选择函数)只会在调用虚拟成员函数时发生,而不是重载非成员函数:
struct Base {
virtual void SomeFunction() {/*do something*/}
};
struct Derived : Base {
virtual void SomeFunction() {/*do something else*/}
};
b->SomeFunction(); // Base version
d->SomeFunction(); // Derived override
bd->SomeFunction(); // Derived override (from dynamic type)
你可以在成员函数的帮助下实现非成员函数的动态调度:
void SomeFunction(Base * b) {b->SomeFunction();}
如评论中所述,此技术可以扩展为实现多个调度,根据多个函数参数的动态类型选择函数。但这超出了问题的范围。