Constructors are called in order of derivation Destructors are called in reverse order in VS .NET

Encode Data Matrix in VS .NET Constructors are called in order of derivation Destructors are called in reverse order

Constructors are called in order of derivation Destructors are called in reverse order
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The results of the foregoing experiment can be generalized as follows: When an object of a derived class is created, the base class constructor is called first, followed by the constructor for the derived class When a derived object is destroyed, its destructor is called first, followed by the destructor for the base class Put differently, constructors are executed in the order of their derivation Destructors are executed in reverse order of derivation If you think about it, it makes sense that constructor functions are executed in the order of their derivation Because a base class has no knowledge of any derived class, any initialization it needs to perform is separate from, and possibly prerequisite to, any initialization performed by the derived class Therefore, it must be executed first Likewise, it is quite sensible that destructors be executed in reverse order of derivation Since the base class underlies a derived class, the destruction of the base class implies the destruction of the derived class Therefore, the derived destructor must be called before the object is fully destroyed In the case of a large class hierarchy (ie, where a derived class becomes the base class for another derived class), the general rule applies: Constructors are called in order of derivation, destructors in reverse order For example, this program
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#include <iostream> using namespace std; class base { public:
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Inheritance
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base() { cout << "Constructing base\n"; } ~base() { cout << "Destructing base\n"; } }; class derived1 : public base { public: derived1() { cout << "Constructing derived1\n"; } ~derived1() { cout << "Destructing derived1\n"; } }; class derived2: public derived1 { public: derived2() { cout << "Constructing derived2\n"; } ~derived2() { cout << "Destructing derived2\n"; } }; int main() { derived2 ob; // construct and destruct ob return 0; }
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displays this output:
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Constructing base Constructing derived1 Constructing derived2 Destructing derived2 Destructing derived1 Destructing base
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The same general rule applies in situations involving multiple base classes For example, this program
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#include <iostream> using namespace std; class base1 { public: base1() { cout << "Constructing base1\n"; } ~base1() { cout << "Destructing base1\n"; } }; class base2 { public: base2() { cout << "Constructing base2\n"; } ~base2() { cout << "Destructing base2\n"; } };
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class derived: public base1, public base2 { public: derived() { cout << "Constructing derived\n"; } ~derived() { cout << "Destructing derived\n"; } }; int main() { derived ob; // construct and destruct ob return 0; }
produces this output:
Constructing base1 Constructing base2 Constructing derived Destructing derived Destructing base2 Destructing base1
As you can see, constructors are called in order of derivation, left to right, as specified in derived s inheritance list Destructors are called in reverse order, right to left This means that if base2 were specified before base1 in derived s list, as shown here:
class derived: public base2, public base1 {
then the output of the preceding program would look like this:
Constructing base2 Constructing base1 Constructing derived Destructing derived Destructing base1 Destructing base2
Passing Parameters to Base Class Constructors
So far, none of the preceding examples have included constructors requiring arguments In cases where only the constructor of the derived class requires one or more arguments, you simply use the standard parameterized constructor syntax But how do you pass arguments to a constructor in a base class The answer is to use an expanded form of the derived class constructor declaration, which passes arguments along to one or more base class constructors The general form of this expanded declaration is shown here:
Inheritance
derived-constructor(arg-list) : base1(arg-list), base2(arg-list), baseN(arg-list); { body of derived constructor } Here, base1 through baseN are the names of the base classes inherited by the derived class Notice that a colon separates the constructor declaration of the derived class from the base classes, and that the base classes are separated from each other by commas, in the case of multiple base classes Consider this sample program:
#include <iostream> using namespace std; class base { protected: int i; public: base(int x) { i = x; cout << "Constructing base\n"; } ~base() { cout << "Destructing base\n"; } }; class derived: public base { int j; public: // derived uses x; y is passed along to base derived(int x, int y): base(y) { j = x; cout << "Constructing derived\n"; } ~derived() { cout << "Destructing derived\n"; } void show() { cout << i << " " << j << "\n"; } }; int main() { derived ob(3, 4); obshow(); return 0; } // displays 4 3
Here, derived s constructor is declared as taking two parameters, x and y However, derived( ) uses only x; y is passed along to base( ) In general, the constructor of the derived class must declare the parameter(s) that its class requires, as well as any required by the base class As the preceding example illustrates, any parameters required by the base class are passed to it in the base class argument list, specified after the colon
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