Ob + 10; // will work in Visual Studio .NET

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Ob + 10; // will work
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Because the object Ob is on the left side of the + operator, it invokes its overloaded operator function, which (presumably) is capable of adding an integer value to some element of Ob However, this statement won t work:
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10 + Ob; // won't work
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The problem with this statement is that the object on the left of the + operator is an integer, a built-in type for which no operation involving an integer and an object of Ob s type is defined The solution to the preceding problem is to overload the + using two friend functions In this case, the operator function is explicitly passed both arguments, and it is invoked like any other overloaded function, based upon the types of its arguments One version of the + operator function handles object + integer, and the other handles integer + object Overloading the + (or any other binary operator) using friend functions allows a built-in type to occur on the left or right side of the operator The following sample program shows you how to accomplish this:
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#include <iostream> using namespace std; class CL {
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public: int count; CL operator=(CL obj); friend CL operator+(CL ob, int i); friend CL operator+(int i, CL ob); }; CL CL::operator=(CL obj) { count = objcount; return *this; } // This handles ob + int CL operator+(CL ob, int i) { CL temp; tempcount = obcount + i; return temp; } // This handles int + ob CL operator+(int i, CL ob) { CL temp; tempcount = obcount + i; return temp; } int main() { CL O; Ocount = 10; cout << Ocount << " "; // outputs 10 O = 10 + O; // add object to integer cout << Ocount << " "; // outputs 20 O = O + 12; // add integer to object cout << Ocount; // outputs 32 return 0; }
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As you can see, the operator+( ) function is overloaded twice, to accommodate the two ways in which an integer and an object of type CL can occur in the addition operation
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You can also overload a unary operator by using a friend function However, doing so requires a little extra effort To begin, think back to the original version of the overloaded ++ operator relative to the three_d class that was implemented as a member function It is shown here for your convenience:
// Overload the prefix form of ++ three_d three_d::operator++() { x++; y++; z++; return *this; }
As you know, every member function receives as an implicit argument this, which is a pointer to the object that invokes the function When a unary operator is overloaded by use of a member function, no argument is explicitly declared The only argument needed in this situation is the implicit pointer to the invoking object Any changes made to the object s data will affect the object on which the operator function is called Therefore, in the preceding function, x++ increments the x member of the invoking object Unlike member functions, a nonmember function, including a friend, does not receive a this pointer, and therefore cannot access the object on which it was called Instead, a friend operator function is passed its operand explicitly For this reason, trying to create a friend operator++( ) function, as shown here, will not work:
// THIS WILL NOT WORK three_d operator++(three_d op1) { op1x++; op1y++; op1z++; return op1; }
This function will not work because only a copy of the object that activated the call to operator++( ) is passed to the function in parameter op1 Thus, the changes inside operator++( ) will not affect the calling object, only the local parameter If you want to use a friend function to overload the increment or decrement operators, you must pass the object to the function as a reference parameter Since a reference parameter is an implicit pointer to the argument, changes to the parameter will affect the argument Using a reference parameter allows the function to increment or decrement the object used as an operand When a friend is used for overloading the increment or decrement operators, the prefix form takes one parameter (which is the operand) The postfix form takes two parameters The second is an integer, which is not used
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