在C++中覆盖类(包含虚析构函数)时，我在继承类中重写了析构函数并将其声明为虚函数，但是我需要调用基类的析构函数吗？
如果需要，我想代码应该是这样的...
MyChildClass::~MyChildClass() // virtual in header
{
// Call to base destructor...
this->MyBaseClass::~MyBaseClass();
// Some destructing specific to MyChildClass
}
我是正确的吗？
不需要手动调用析构函数，析构函数会自动按照构造函数的相反顺序被调用（基类的析构函数最后被调用）。不要手动调用基类的析构函数。
class B
{
public:
virtual ~B()
{
cout<<"B destructor"<<endl;
}
};
class D : public B
{
public:
virtual ~D()
{
cout<<"D destructor"<<endl;
}
};
当你执行以下操作时：
B *pD = new D();
delete pD;
如果B类中没有虚析构函数，只有~B()会被调用。但是由于你在D类中声明了虚析构函数，所以首先会调用~D()，然后调用~B()。
其他人说的都对，但也要注意在派生类中不必声明析构函数为虚函数。一旦你在基类中声明了析构函数为虚函数，所有派生类的析构函数都将是虚函数，无论你是否声明它们为虚函数。换句话说：
struct A {
virtual ~A() {}
};
struct B : public A {
virtual ~B() {} // this is virtual
};
struct C : public A {
~C() {} // this is virtual too
};
C++中的析构函数会在它们被构造的顺序（派生类在基类之后）中自动调用，但仅当基类的析构函数被声明为virtual时。
如果没有声明为virtual，则在对象删除时只会调用基类的析构函数。
例如：没有虚拟析构函数的情况
#include <iostream>
using namespace std;
class Base{
public:
Base(){
cout << "Base Constructor \n";
}
~Base(){
cout << "Base Destructor \n";
}
};
class Derived: public Base{
public:
int *n;
Derived(){
cout << "Derived Constructor \n";
n = new int(10);
}
void display(){
cout<< "Value: "<< *n << endl;
}
~Derived(){
cout << "Derived Destructor \n";
}
};
int main() {
Base *obj = new Derived(); //Derived object with base pointer
delete(obj); //Deleting object
return 0;
}
输出
Base Constructor
Derived Constructor
Base Destructor
示例：使用基类虚析构函数
#include <iostream>
using namespace std;
class Base{
public:
Base(){
cout << "Base Constructor \n";
}
//virtual destructor
virtual ~Base(){
cout << "Base Destructor \n";
}
};
class Derived: public Base{
public:
int *n;
Derived(){
cout << "Derived Constructor \n";
n = new int(10);
}
void display(){
cout<< "Value: "<< *n << endl;
}
~Derived(){
cout << "Derived Destructor \n";
delete(n); //deleting the memory used by pointer
}
};
int main() {
Base *obj = new Derived(); //Derived object with base pointer
delete(obj); //Deleting object
return 0;
}
输出
Base Constructor
Derived Constructor
Derived Destructor
Base Destructor
建议将基类析构函数声明为virtual，否则会导致未定义的行为。
参考：虚拟析构函数
不，你从来没有调用基类的析构函数，正如其他人指出的那样，它总是自动调用的，但这里有一个概念证明和结果：
class base {
public:
base() { cout << __FUNCTION__ << endl; }
~base() { cout << __FUNCTION__ << endl; }
};
class derived : public base {
public:
derived() { cout << __FUNCTION__ << endl; }
~derived() { cout << __FUNCTION__ << endl; } // adding call to base::~base() here results in double call to base destructor
};
int main()
{
cout << "case 1, declared as local variable on stack" << endl << endl;
{
derived d1;
}
cout << endl << endl;
cout << "case 2, created using new, assigned to derive class" << endl << endl;
derived * d2 = new derived;
delete d2;
cout << endl << endl;
cout << "case 3, created with new, assigned to base class" << endl << endl;
base * d3 = new derived;
delete d3;
cout << endl;
return 0;
}
输出结果为：
case 1, declared as local variable on stack
base::base
derived::derived
derived::~derived
base::~base
case 2, created using new, assigned to derive class
base::base
derived::derived
derived::~derived
base::~base
case 3, created with new, assigned to base class
base::base
derived::derived
base::~base
Press any key to continue . . .
如果将基类的析构函数声明为虚函数，那么情况3的结果将与情况1和2相同。
不同于其他虚方法，在这里，您不需要显式地从派生类调用基类方法来“链接”调用，编译器会生成代码以按照构造函数调用的相反顺序调用析构函数。
delete两次会导致分段错误？ - Géry Ogam