C++ Primer Plus

1. Both versions give the same answers, but the if else version is more efficient. Consider what happens, for example, when ch is a space. Version 1, after incrementing spaces, tests whether the character is a newline. This wastes time because the program has already established that ch is a space and hence could not be a newline. Version 2, in the same situation, skips the newline test.

2. Both ++ch and ch + 1 have the same numerical value. But ++ch is type char and prints as a character, while ch + 1, because it adds a char to an int, is type int and prints as a number.

3. Because the program uses ch = '$' instead of ch == '$', the combined input and output looks like this:

Hi!

H$i$!$

$Send $10 or $20 now!

S$e$n$d$ $ct1 = 9, ct2 = 9

Each character is converted to the $ character before being printed the second time. Also the value of the expression ch = $ is the code for the $ character, hence nonzero, hence true; so ct2 is incremented each time.

4.

a. weight >= 115 && weight < 125

b. ch == 'q' || ch == 'Q'

c. x % 2 == 0 && x != 26

d. x % 2 == 0 && !(x % 26 == 0)

e. donation >= 1000 && donation <= 2000 || guest == 1

f. (ch >= 'a' && ch <= 'z') ||(ch >= 'A' && ch <= 'Z')

5. Not necessarily. For example, if x is 10, then !x is 0 and !!x is 1. However, if x is a bool variable, then !!x is x.

6. (x < 0)? -x : x

or

(x >= 0)? x : -x;

7.

switch (ch)

{

    case 'A': a_grade++;

              break;

    case 'B': b_grade++;

              break;

    case 'C': c_grade++;

              break;

    case 'D': d_grade++;

              break;

    default:  f_grade++;

              break;

}

8. If you use integer labels and the user types a noninteger such as q, the program hangs because integer input can’t process a character. But if you use character labels and the user types an integer such as 5, character input will process 5 as a character. Then the default part of the switch can suggest entering another character.

9. Here is one version:

int line = 0;

char ch;

while (cin.get(ch) && ch != 'Q')

{

    if (ch == '\n')

        line++;

}

Answers to Chapter Review for Chapter 7

1. The three steps are defining the function, providing a prototype, and calling the function.

2.

a. void igor(void); // or void igor()

b. float tofu(int n); // or float tofu(int);

c. double mpg(double miles, double gallons);

d. long summation(long harray[], int size);

e. double doctor(const char * str);

f. void ofcourse(boss dude);

g. char * plot(map *pmap);

3.

void set_array(int arr[], int size, int value)

{

    for (int i = 0; i < size; i++)

        arr[i] = value;

}

4.

void set_array(int * begin, int * end, int value)

{

    for (int * pt = begin; pt != end; pt++)

        pt* = value;

}

5.

double biggest (const double foot[], int size)

{

    double max;

    if (size < 1)

    {

        cout << "Invalid array size of " << size << endl;

        cout << "Returning a value of 0\n";

        return 0;

    }

    else    // not necessary because return terminates program

    {

        max = foot[0];

        for (int i = 1; i < size; i++)

            if (foot[i] > max)

                max = foot[i];

        return max;

    }

}

6. You use the const qualifier with pointers to protect the original pointed-to data from being altered. When a program passes a fundamental type such as an int or a double, it passes it by value so that the function works with a copy. Thus, the original data is already protected.

7. A string can be stored in a char array, it can be represented by a string constant in double quotation marks, and it can be represented by a pointer pointing to the first character of a string.

8.

int replace(char * str, char c1, char c2)

{

    int count = 0;

    while (*str)      // while not at end of string

    {

        if (*str == c1)

        {

            *str = c2;

            count++;

        }

        str++;        // advance to next character

    }

    return count;

}

9. Because C++ interprets "pizza" as the address of its first element, applying the * operator yields the value of that first element, which is the character p. Because C++ interprets "taco" as the address of its first element, it interprets "taco"[2] as the value of the element two positions down the line—that is, as the character c. In other words, the string constant acts the same as an array name.

10. To pass it by value, you just pass the structure name glitz. To pass its address, you use the address operator &glitz. Passing by the value automatically protects the original data, but it takes time and memory. Passing by address saves time and memory but doesn’t protect the original data unless you use the const modifier for the function parameter. Also passing by value means you can use ordinary structure member notation, but passing a pointer means you have to remember to use the indirect membership operator.

11. int judge (int (*pf)(const char *));

12.

a. Note that if ap is an applicant structure, then ap.credit_ratings is an array name and ap.credit_ratings[i] is an array element.

void display(applicant ap)

{

    cout << ap.name << endl;

    for (int i = 0; i < 3; i++)

        cout  << ap.credit_ratings[i] <<  endl;

}

b. Note that if pa is a pointer to an applicant structure, then pa->credit_ratings is an array name and pa->credit_ratings[i] is an array element.

void show(const applicant * pa)

{