我对大多数OOP理论都有很好的理解,但最让我困惑的是虚拟析构函数。
我以为析构函数总是被调用,不管是什么,也不管是链中的每个对象。
你打算什么时候让它们虚拟化?为什么?
我对大多数OOP理论都有很好的理解,但最让我困惑的是虚拟析构函数。
我以为析构函数总是被调用,不管是什么,也不管是链中的每个对象。
你打算什么时候让它们虚拟化?为什么?
当前回答
简单地说,当您删除指向派生类对象的基类指针时,虚拟析构函数将以正确的顺序析构函数资源。
#include<iostream>
using namespace std;
class B{
public:
B(){
cout<<"B()\n";
}
virtual ~B(){
cout<<"~B()\n";
}
};
class D: public B{
public:
D(){
cout<<"D()\n";
}
~D(){
cout<<"~D()\n";
}
};
int main(){
B *b = new D();
delete b;
return 0;
}
OUTPUT:
B()
D()
~D()
~B()
==============
If you don't give ~B() as virtual. then output would be
B()
D()
~B()
where destruction of ~D() is not done which leads to leak
其他回答
虚拟构造函数是不可能的,但虚拟析构函数是可能的。让我们做个实验。。。。。。。
#include <iostream>
using namespace std;
class Base
{
public:
Base(){
cout << "Base Constructor Called\n";
}
~Base(){
cout << "Base Destructor called\n";
}
};
class Derived1: public Base
{
public:
Derived1(){
cout << "Derived constructor called\n";
}
~Derived1(){
cout << "Derived destructor called\n";
}
};
int main()
{
Base *b = new Derived1();
delete b;
}
上述代码输出以下内容:
Base Constructor Called
Derived constructor called
Base Destructor called
派生对象的构造遵循构造规则,但当我们删除“b”指针(基指针)时,我们发现只有基析构函数被调用。但这绝不能发生。为了做适当的事情,我们必须使基析构函数虚拟化。现在让我们看看以下情况:
#include <iostream>
using namespace std;
class Base
{
public:
Base(){
cout << "Base Constructor Called\n";
}
virtual ~Base(){
cout << "Base Destructor called\n";
}
};
class Derived1: public Base
{
public:
Derived1(){
cout << "Derived constructor called\n";
}
~Derived1(){
cout << "Derived destructor called\n";
}
};
int main()
{
Base *b = new Derived1();
delete b;
}
输出变化如下:
Base Constructor Called
Derived Constructor called
Derived destructor called
Base destructor called
因此,基指针的销毁(对派生对象进行分配!)遵循销毁规则,即首先是派生指针,然后是基指针。另一方面,没有什么像虚拟构造函数。
只要类是多态的,就将析构函数设为虚拟。
当您希望不同的析构函数在通过基类指针删除对象时遵循正确的顺序时,析构函数的虚拟关键字是必需的。例如:
Base *myObj = new Derived();
// Some code which is using myObj object
myObj->fun();
//Now delete the object
delete myObj ;
如果基类析构函数是虚拟的,那么对象将按顺序被析构函数(首先是派生对象,然后是基)。如果基类析构函数不是虚拟的,那么只有基类对象会被删除(因为指针是基类“base*myObj”)。因此派生对象将存在内存泄漏。
如果使用shared_ptr(仅shared_ptl,而不是unique_ptr),则不必将基类析构函数设为虚拟:
#include <iostream>
#include <memory>
using namespace std;
class Base
{
public:
Base(){
cout << "Base Constructor Called\n";
}
~Base(){ // not virtual
cout << "Base Destructor called\n";
}
};
class Derived: public Base
{
public:
Derived(){
cout << "Derived constructor called\n";
}
~Derived(){
cout << "Derived destructor called\n";
}
};
int main()
{
shared_ptr<Base> b(new Derived());
}
输出:
Base Constructor Called
Derived constructor called
Derived destructor called
Base Destructor called
我认为这个问题的核心是关于虚拟方法和多态性,而不是具体的析构函数。下面是一个更清晰的例子:
class A
{
public:
A() {}
virtual void foo()
{
cout << "This is A." << endl;
}
};
class B : public A
{
public:
B() {}
void foo()
{
cout << "This is B." << endl;
}
};
int main(int argc, char* argv[])
{
A *a = new B();
a->foo();
if(a != NULL)
delete a;
return 0;
}
将打印出:
This is B.
如果没有虚拟,它将打印出:
This is A.
现在您应该了解何时使用虚拟析构函数。