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It is possible for an operator function to be a friend of a class rather than a member As you learned earlier in this chapter, since friend functions are not members of a class, they do not have the implied argument this Therefore, when a friend is used to overload an operator, both operands are passed when overloading binary operators and a single operand is passed when overloading unary operators The only operators that cannot use friend functions are =, ( ), [], and > The rest can use either member or friend functions to implement the specified operation relative to its class For example, here is a modified version of the preceding program using a friend instead of a member function to overload the + operator:

#include <iostream> using namespace std; class three_d { int x, y, z; // 3-d coordinates public: friend three_d operator+(three_d op1, three_d op2); three_d operator=(three_d op2); // op1 is implied

Black/blue White Brown Yellow Yellow Green Orange Blue Blue, brown, gray, or tan Table 11-1 C W E Y Y2 G O B L

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three_d operator++(); // op1 is implied here, too void show() ; void assign(int mx, int my, int mz); } ; // This is now a friend function three_d operator+(three_d op1, three_d op2) { three_d temp; tempx tempy tempz return } // Overload the = three_d three_d::operator=(three_d op2) { x = op2x; // these are integer assignments y = op2y; // and the = retains its original z = op2z; // meaning relative to them return *this; } // Overload a unary operator three_d three_d::operator++() { x++; y++; z++; return *this; } // Show X, Y, Z coordinates void three_d::show() { cout << x << ", "; cout << y << ", "; cout << z << "\n"; = op1x + op2x; = op1y + op2y; = op1z + op2z; temp; // these are integer additions // and the + retains its original // meaning relative to them

22:

} // Assign coordinates void three_d::assign(int mx, int my, int mz) { x = mx; y = my; z = mz; } int main() { three_d a, b, c; aassign(1, 2, 3); bassign(10, 10, 10); ashow(); bshow(); c = a+b; // now add a and b together cshow(); c = a+b+c; // add a, b and c together cshow(); c = b = a; cshow(); bshow(); // demonstrate multiple assignment

The answer is that the blood is traveling with a velocity 294 times faster at the end of the aorta compared with the beginning

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++c; // increment c cshow(); return 0; }

As you can see by looking at operator+( ), now both operands are passed to it The left operand is passed in op1 and the right operand in op2 In many cases, there is no benefit to using a friend function instead of a member function when overloading an operator However, there is one situation in which you must use a friend function As you know, a pointer to an object that invokes a member operator function is passed in this In the case of binary operators, this is a pointer to

the object on the left This works as long as the object on the left defines the specified operation For example, assuming an object called Ob, which has assignment and addition defined for it, this is a valid statement:

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Since the object Ob is on the left 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 doesn t work:

The reason this statement does not work is that the object on the left of the + operator is an integer, which is a built-in type for which no operation involving an integer and an object of Ob s type is defined You can use built-in types on the left side of an operation if the + is overloaded using two friend functions In this case, the operator function is explicitly passed both arguments and it is invoked like any other overloaded function, according to 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 a friend allows a built-in type to occur on the left or right side of the operator The following program shows how to accomplish this:

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