C++ Primer Plus

In Listing 3.11, the division operator represents three distinct operations: int division, float division, and double division. C++ uses the context—in this case the type of operands—to determine which operator is meant. The process of using the same symbol for more than one operation is called operator overloading. C++ has a few examples of overloading built in to the language. C++ also lets you extend operator overloading to user-defined classes, so what you see here is a precursor of an important OOP property (see Figure 3.4).

Figure 3.4. Different divisions.

The Modulus Operator

Most people are more familiar with addition, subtraction, multiplication, and division than with the modulus operation, so let’s take a moment to look at the modulus operator in action. The modulus operator returns the remainder of an integer division. In combination with integer division, the modulus operation is particularly useful in problems that require dividing a quantity into different integral units, such as converting inches to feet and inches or converting dollars to quarters, dimes, nickels, and pennies. In Chapter 2, Listing 2.6 converts weight in British stone to pounds. Listing 3.12 reverses the process, converting weight in pounds to stone. A stone, you remember, is 14 pounds, and most British bathroom scales are calibrated in this unit. The program uses integer division to find the largest number of whole stone in the weight, and it uses the modulus operator to find the number of pounds left over.

Listing 3.12. modulus.cpp

// modulus.cpp -- uses % operator to convert lbs to stone

#include

int main()

{

    using namespace std;

    const int Lbs_per_stn = 14;

    int lbs;

    cout << "Enter your weight in pounds: ";

    cin >> lbs;

    int stone = lbs / Lbs_per_stn;      // whole stone

    int pounds = lbs % Lbs_per_stn;     // remainder in pounds

    cout << lbs << " pounds are " << stone

         << " stone, " << pounds << " pound(s).\n";

    return 0;

}

Here is a sample run of the program in Listing 3.12:

Enter your weight in pounds: 181

181 pounds are 12 stone, 13 pound(s).

In the expression lbs / Lbs_per_stn, both operands are type int, so the computer performs integer division. With a lbs value of 181, the expression evaluates to 12. The product of 12 and 14 is 168, so the remainder of dividing 14 into 181 is 13, and that’s the value of lbs % Lbs_per_stn. Now you are prepared technically, if not emotionally, to respond to questions about your weight when you travel in Great Britain.

Type Conversions

C++’s profusion of types lets you match the type to the need. It also complicates life for the computer. For example, adding two short values may involve different hardware instructions than adding two long values. With 11 integer types and 3 floating-point types, the computer can have a lot of different cases to handle, especially if you start mixing types. To help deal with this potential mishmash, C++ makes many type conversions automatically:

• C++ converts values when you assign a value of one arithmetic type to a variable of another arithmetic type.

• C++ converts values when you combine mixed types in expressions.

• C++ converts values when you pass arguments to functions.

If you don’t understand what happens in these automatic conversions, you might find some program results baffling, so let’s take a more detailed look at the rules.

Conversion on Initialization and Assignment

C++ is fairly liberal in allowing you to assign a numeric value of one type to a variable of another type. Whenever you do so, the value is converted to the type of the receiving variable. For example, suppose so_long is type long, thirty is type short, and you have the following statement in a program:

so_long = thirty;            // assigning a short to a long

The program takes the value of thirty (typically a 16-bit value) and expands it to a long value (typically a 32-bit value) upon making the assignment. Note that the expansion creates a new value to place into so_long; the contents of thirty are unaltered.

Assigning a value to a type with a greater range usually poses no problem. For example, assigning a short value to a long variable doesn’t change the value; it just gives the value a few more bytes in which to laze about. However, assigning a large long value such as 2111222333 to a float variable results in the loss of some precision. Because float can have just six significant figures, the value can be rounded to 2.11122E9. So while some conversions are safe, some may pose difficulties. Table 3.3 points out some possible conversion problems.

Table 3.3. Potential Numeric Conversion Problems

A zero value assigned to a bool variable is converted to false, and a nonzero value is converted to true.