一、相關問題：
1. 基類、派生類的構造和析構順序
2. 基類、派生類中virtual的取舍
二、測試代碼：
#include <iostream>
class A {
public:
A() {
std::cout << "A()" << std::endl;
}
virtual void print() { std::cout << "A print()" << std::endl; }
virtual ~A() {
std::cout << "~A()" << std::endl;
}
};
class B : public A {
public:
B() : A(){
std::cout << "B()" << std::endl;
}
virtual void print() { std::cout << "B print()" << std::endl; }
virtual ~B() {
std::cout << "~B()" << std::endl;
}
};
void print() {
}
int main(int argc, char *argv[])
{
A* c = new B();
c->print();
delete c;
return 0;
}
三、探討與結(jié)論：
1. 基類、派生類的構造和析構順序為：基類構造-派生類構造-派生類析構-基類析構
上述代碼輸出結(jié)果為：
2. 基類、派生類中virtual的取舍：若要實現(xiàn)動態(tài)綁定，基類中virtual關鍵字不可舍棄，派生類中virtual關鍵字可有可無；若基類中有關鍵字virtual，則普通函數(shù)調(diào)用派生類函數(shù)，析構函數(shù)先調(diào)用派生類，再調(diào)用基類；若基類中無關鍵字virtual，則普通函數(shù)和析構函數(shù)均只調(diào)用基類函數(shù)。
測試代碼1：基類無關鍵字virtual
#include <iostream>
class A {
public:
A() {
std::cout << "A()" << std::endl;
}
void print() { std::cout << "A print()" << std::endl; }
~A() {
std::cout << "~A()" << std::endl;
}
};
class B : public A {
public:
B() : A(){
std::cout << "B()" << std::endl;
}
virtual void print() { std::cout << "B print()" << std::endl; }
virtual ~B() {
std::cout << "~B()" << std::endl;
}
};
void print() {
}
int main(int argc, char *argv[])
{
A* c = new B();
c->print();
delete c;
return 0;
}
輸出結(jié)果為：
測試代碼2：派生類無關鍵字virtual
#include <iostream>
class A {
public:
A() {
std::cout << "A()" << std::endl;
}
virtual void print() { std::cout << "A print()" << std::endl; }
virtual ~A() {
std::cout << "~A()" << std::endl;
}
};
class B : public A {
public:
B() : A(){
std::cout << "B()" << std::endl;
}
void print() { std::cout << "B print()" << std::endl; }
~B() {
std::cout << "~B()" << std::endl;
}
};
void print() {
}
int main(int argc, char *argv[])
{
A* c = new B();
c->print();
delete c;
return 0;
}
輸出結(jié)果為：
