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Tip of the Day
Home » Tip Bank » C++
Language: C
Expertise: Beginner
Nov 20, 2003

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Application Security Testing: An Integral Part of DevOps


Make an Object Appear Like a Pointer

You can make an object appear like a pointer with a smart pointer. If a class overloads the operator ->( ), then any object of that class can appear like a pointer when the operator ->( ) is called. The following program illustrates this:

#include "iostream.h"
class test
{
public :
void fun( )
{
cout << "fun of smart pointer" ;
}
} ;

class smartpointer
{
test t ;
public :
test* operator ->( )
{
return &t ;
}
} ;

void main( )
{
smartpointer sp ;
sp -> fun( ) ;
}
The beauty of overloading operator ->( ) is that even though sp is an object, you can make it work like a pointer. The operator ->( ) returns the address of the object of the type test. Using this address of the test object the function fun( ) of the class test gets called. Thus even though fun( ) is not a member of smartpointer class, you can still call it using sp.
Murali Krishna
 
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