C++ Primer Plus

Program Notes

Normally cout drops trailing zeros. For example, it would display 3333333.250000 as 3333333.25. The call to cout.setf() overrides that behavior, at least in new implementations. The main thing to note in Listing 3.8 is how float has less precision than double. Both tub and mint are initialized to 10.0 / 3.0. That should evaluate to 3.33333333333333333...(etc.). Because cout prints six figures to the right of the decimal, you can see that both tub and mint are accurate that far. But after the program multiplies each number by a million, you see that tub diverges from the proper value after the seventh three. tub is good to seven significant figures. (This system guarantees six significant figures for float, but that’s the worst-case scenario.) The type double variable, however, shows 13 threes, so it’s good to at least 13 significant figures. Because the system guarantees 15, this shouldn’t surprise you. Also note that multiplying a million tubs by 10 doesn’t quite result in the correct answer; this again points out the limitations of float precision.

The ostream class to which cout belongs has class member functions that give you precise control over how the output is formatted—field widths, places to the right of the decimal point, decimal form or E form, and so on. Chapter 17 outlines those choices. This book’s examples keep it simple and usually just use the << operator. Occasionally, this practice displays more digits than necessary, but that causes only aesthetic harm. If you do mind, you can skim Chapter 17 to see how to use the formatting methods. Don’t, however, expect to fully follow the explanations at this point.

Reading Include Files

The include directives found at the top of C++ source files often take on the air of a magical incantation; novice C++ programmers learn, through reading and experience, which header files add particular functionalities, and they include them solely to make their programs work. Don’t rely on the include files only as a source of mystic and arcane knowledge; feel free to open them up and read them. They are text files, so you can read them easily. All the files you include in your programs exist on your computer or in a place where your computer can use them. Find the includes you use and see what they contain. You’ll quickly see that the source and header files you use are an excellent source of knowledge and information—in some cases, the best documentation available. Later, as you progress into more complex inclusions and begin to use other, nonstandard libraries in your applications, this habit will serve you well.

Floating-Point Constants

When you write a floating-point constant in a program, in which floating-point type does the program store it? By default, floating-point constants such as 8.24 and 2.4E8 are type double. If you want a constant to be type float, you use an f or F suffix. For type long double, you use an l or L suffix. (Because the lowercase l looks a lot like the digit 1, the uppercase L is a better choice.) Here are some samples:

1.234f         // a float constant

2.45E20F       // a float constant

2.345324E28    // a double constant

2.2L           // a long double constant

Advantages and Disadvantages of Floating-Point Numbers

Floating-point numbers have two advantages over integers. First, they can represent values between integers. Second, because of the scaling factor, they can represent a much greater range of values. On the other hand, floating point operations usually are slightly slower than integer operations, and you can lose precision. Listing 3.9 illustrates the last point.

Listing 3.9. fltadd.cpp

// fltadd.cpp -- precision problems with float

#include

int main()

{

    using namespace std;

    float a = 2.34E+22f;

    float b = a + 1.0f;

    cout << "a = " << a << endl;

    cout << "b - a = " << b - a << endl;

    return 0;

}

The program in Listing 3.9 takes a number, adds 1, and then subtracts the original number. That should result in a value of 1. Does it? Here is the output from the program in Listing 3.9 for one system:

a = 2.34e+022

b - a = 0

The problem is that 2.34E+22 represents a number with 23 digits to the left of the decimal. By adding 1, you are attempting to add 1 to the 23rd digit in that number. But type float can represent only the first 6 or 7 digits in a number, so trying to change the 23rd digit has no effect on the value.

Classifying Data Types