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Recipe 8.5. Ensuring a Single Copy of a Member Variable
Problem
You have a member variable that you want only one instance of, no matter how many instances of the
class are created. This kind of member variable is generally called a static member or a class variable, as
opposed to an instance variable, which is one that is instantiated with every object of a class.
Solution
Declare the member variable with the static keyword, then initialize it in a separate source file (not the
header file where you declared it) as in Example 8-5
.
Example 8-5. Using a static member variable
// Static.h
class OneStatic {
public:
int getCount( ) {return count;}
OneStatic( );
protected:
static int count;
};
// Static.cpp
#include "Static.h"
int OneStatic::count = 0;
OneStatic::OneStatic( ) {
count++;
}
// StaticMain.cpp
#include <iostream>
#include "static.h"
using namespace std;
int main( ) {
OneStatic a;
OneStatic b;
OneStatic c;
cout << a.getCount( ) << endl;
cout << b.getCount( ) << endl;
cout << c.getCount( ) << endl;
}
Discussion
static is C++'s way of allowing only one copy of something. If you declare a member variable static, only
one of it will ever be constructed, regardless of the number of objects of that class that are instantiated.
Similarly, if you declare a variable static in a function, it is constructed at most once and retains its value
from one function call to another. With member variables, you have to do a little extra work to make sure
member variables are allocated properly, though. This is why there are three files in Example 8-5
.
First, you have to use the static keyword when you declare the variable. This is easy enough: add this
keyword in the class header in the header file Static.h:
protected:
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Recipe 8.6. Determining an Object's Type at Runtime
Problem
At runtime, you need to interrogate dynamically the type of particular class.
Solution
Use runtime type identification (commonly referred to as RTTI) to query the address of the object for
the type of object it points to. Example 8-6
shows how.
Example 8-6. Using runtime type identification
#include <iostream>
#include <typeinfo>
using namespace std;
class Base {};
class Derived : public Base {};
int main( ) {
Base b, bb;
Derived d;
// Use typeid to test type equality
if (typeid(b) == typeid(d)) { // No
cout << "b and d are of the same type.\n";
}
if (typeid(b) == typeid(bb)) { // Yes
cout << "b and bb are of the same type.\n";
}
if (typeid(d) == typeid(Derived)) { // Yes
cout << "d is of type Derived.\n";
}
}
Discussion
Example 8-6
shows you how to use the operator typeid to determine and compare the type of an
object. typeid takes an expression or a type and returns a reference to an object of type_info or a
subclass of it (which is implementation defined). You can use what is returned to test for equality or
retrieve a string representation of the type's name. For example, you can compare the types of two
objects like this:
if (typeid(b) == typeid(d)) {
This will return true if the type_info objects returned by both of these are equal. This is because typeid
returns a reference to a static object, so if you call it on two objects that are the same type, you will get
two references to the same thing, which is why the equality test returns true.
You can also use typeid with the type itself, as in:
if (typeid(d) == typeid(Derived)) {
This allows you to explicitly test for a particular type.
Probably the most common use of typeid is for debugging. To write out the name of the type, use
type_info::name, like this:
std::cout << typeid(d).name( ) << std::endl;
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Recipe 8.7. Determining if One Object's Class Is a Subclass of
Another
Problem
You have two objects, and you need to know if their respective classes have a base class/derived class
relationship or if they are unrelated.
Solution
Use the dynamic_cast operator to attempt to downcast from one type to another. The result tells you
about the class's relationships. Example 8-7
presents some code for doing this.
Example 8-7. Determining class relationships
#include <iostream>
#include <typeinfo>
using namespace std;
class Base {
public:
virtual ~Base( ) {} // Make this a polymorphic class
};
class Derived : public Base {
public:
virtual ~Derived( ) {}
};
int main( ) {
Derived d;
// Query the type relationship
if (dynamic_cast<Base*>(&d)) {
cout << "Derived is a subclass of Base" << endl;
}
else {
cout << "Derived is NOT a subclass of Base" << endl;
}
}
Discussion
Use the dynamic_cast operator to query the relationship between two types. dynamic_cast takes a
pointer or reference to a given type and tries to convert it to a pointer or reference of a derived type, i.e.,
casting down a class hierarchy. If you have a Base* that points to a Derived object,
dynamic_cast<Base*>(&d) returns a pointer of type Derived only if d is an object of a type that's
derived from Base. If this is not possible (because Derived is not a subclass, directly or indirectly, of
Base), the cast fails and NULL is returned if you passed dynamic_cast a pointer to a derived object. If it
is a reference, then the standard exception bad_cast is thrown. Also, the base class must be publicly
inherited and it must be unambiguous. The result tells you if one class is a descendant of another. Here's
what I did in Example 8-7
:
if (dynamic_cast<Base*>(&d)) {
This returns a non-NULL pointer because d is an object of a class that is a descendant of Base. Use this
on any pair of classes to determine their relationship. The only requirement is that the object argument is a
polymorphic type, which means that it has at least one virtual function. If it does not, it won't compile.
This doesn't usually cause much of a headache though, because a class hierarchy without virtual functions
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