Adding an Addition Operator

It’s a simple matter to convert the Time class to using an overloaded addition operator. You just change the name of Sum() to the odder-looking name operator+(). That’s right: You just append the operator symbol (+, in this case) to the end of operator and use the result as a method name. This is one place where you can use a character other than a letter, a digit, or an underscore in an identifier name. Listings 11.4 and 11.5 reflect this small change.

Listing 11.4. mytime1.h

// mytime1.h -- Time class before operator overloading
#ifndef MYTIME1_H_
#define MYTIME1_H_

class Time
{
private:
    int hours;
    int minutes;
public:
    Time();
    Time(int h, int m = 0);
    void AddMin(int m);
    void AddHr(int h);
    void Reset(int h = 0, int m = 0);
    Time operator+(const Time & t) const;
    void Show() const;
};
#endif

Listing 11.5. mytime1.cpp

// mytime1.cpp  -- implementing Time methods
#include <iostream>
#include "mytime1.h"

Time::Time()
{
    hours = minutes = 0;
}

Time::Time(int h, int m )
{
    hours = h;
    minutes = m;
}

void Time::AddMin(int m)
{
    minutes += m;
    hours += minutes / 60;
    minutes %= 60;
}

void Time::AddHr(int h)
{
    hours += h;
}

void Time::Reset(int h, int m)
{
    hours = h;
    minutes = m;
}

Time Time::operator+(const Time & t) const
{
    Time sum;
    sum.minutes = minutes + t.minutes;
    sum.hours = hours + t.hours + sum.minutes / 60;
    sum.minutes %= 60;
    return sum;
}

void Time::Show() const
{
    std::cout << hours << " hours, " << minutes << " minutes";
}

Like Sum(), operator+() is invoked by a Time object, takes a second Time object as an argument, and returns a Time object. Thus, you can invoke the operator+() method by using the same syntax that Sum() uses:

total = coding.operator+(fixing);    // function notation

But naming the method operator+() also lets you use operator notation:

total = coding + fixing;              // operator notation

Either notation invokes the operator+() method. Note that with the operator notation, the object to the left of the operator (coding, in this case) is the invoking object, and the object to the right (fixing, in this case) is the one passed as an argument. Listing 11.6 illustrates this point.

Listing 11.6. usetime1.cpp

// usetime1.cpp -- using the second draft of the Time class
// compile usetime1.cpp and mytime1.cpp together
#include <iostream>
#include "mytime1.h"

int main()
{
    using std::cout;
    using std::endl;
    Time planning;
    Time coding(2, 40);
    Time fixing(5, 55);
    Time total;

    cout << "planning time = ";
    planning.Show();
    cout << endl;

    cout << "coding time = ";
    coding.Show();
    cout << endl;

    cout << "fixing time = ";
    fixing.Show();
    cout << endl;

    total = coding + fixing;
    // operator notation
    cout << "coding + fixing = ";
    total.Show();
    cout << endl;

    Time morefixing(3, 28);
    cout << "more fixing time = ";
    morefixing.Show();
    cout << endl;
    total = morefixing.operator+(total);
    // function notation
    cout << "morefixing.operator+(total) = ";
    total.Show();
    cout << endl;

    return 0;
}

Here is the output of the program in Listings 11.4, 11.5, and 11.6:

planning time = 0 hours, 0 minutes
coding time = 2 hours, 40 minutes
fixing time = 5 hours, 55 minutes
coding + fixing = 8 hours, 35 minutes
more fixing time = 3 hours, 28 minutes
morefixing.operator+(total) = 12 hours, 3 minutes

In short, the name of the operator+() function allows it to be invoked by using either function notation or operator notation. The compiler uses the operand types to figure out what to do:

int a, b, c;
Time A, B, C;
c = a + b;      // use int addition
C = A + B;      // use addition as defined for Time objects

Can you add more than two objects? For example, if t1, t2, t3, and t4 are all Time objects, can you do the following?

t4 = t1 + t2 + t3;                     // valid?

The way to answer this is to consider how the statement gets translated into function calls. Because addition is a left-to-right operator, the statement is first translated to this:

t4 = t1.operator+(t2 + t3);            // valid?

Then the function argument is itself translated to a function call, giving the following:

t4 = t1.operator+(t2.operator+(t3));   // valid? YES

Is this valid? Yes, it is. The function call t2.operator+(t3) returns a Time object that represents the sum of t2 and t3. This object then becomes the object of the t1.operator+() function call, and that call returns the sum of t1 and the Time object that represents the sum of t2 and t3. In short, the final return value is the sum of t1, t2, and t3, just as desired.