6、tdio.h>
int main(void)
{
float original_amount, amount_with_tax;
printf("Enter an amount: ");
scanf("%f", &original_amount);
amount_with_tax = original_amount * 1.05f;
printf("With tax added: $%.2f\n", amount_with_tax);
return 0;
}
The amount_with_tax variable is unnecess
7、ary. If we remove it, the program is slightly shorter:
#include
int main(void)
{
float original_amount;
printf("Enter an amount: ");
scanf("%f", &original_amount);
printf("With tax added: $%.2f\n", original_amount * 1.05f);
return 0;
}
Chapter 3
Answers to Sele
8、cted Exercises
2. [was #2]
(a) printf("%-8.1e", x);
(b) printf("%10.6e", x);
(c) printf("%-8.3f", x);
(d) printf("%6.0f", x);
5. [was #8] The values of x, i, and y will be 12.3, 45, and .6, respectively.
Answers to Selected Programming Projects
1. [was #4; modified]
#include
9、
int main(void)
{
int month, day, year;
printf("Enter a date (mm/dd/yyyy): ");
scanf("%d/%d/%d", &month, &day, &year);
printf("You entered the date %d%.2d%.2d\n", year, month, day);
return 0;
}
3. [was #6; modified]
#include
int main(void)
{
int prefix, gr
10、oup, publisher, item, check_digit;
printf("Enter ISBN: ");
scanf("%d-%d-%d-%d-%d", &prefix, &group, &publisher, &item, &check_digit);
printf("GS1 prefix: %d\n", prefix);
printf("Group identifier: %d\n", group);
printf("Publisher code: %d\n", publisher);
printf("Item number: %d\
11、n", item);
printf("Check digit: %d\n", check_digit);
/* The five printf calls can be combined as follows:
printf("GS1 prefix: %d\nGroup identifier: %d\nPublisher code: %d\nItem number: %d\nCheck digit: %d\n",
prefix, group, publisher, item, check_digit);
*/
retur
12、n 0;
}
Chapter 4
Answers to Selected Exercises
2. [was #2] Not in C89. Suppose that i is 9 and j is 7. The value of (-i)/j could be either –1 or –2, depending on the implementation. On the other hand, the value of -(i/j) is always –1, regardless of the implementation. In C99, on the other hand
13、, the value of (-i)/j must be equal to the value of -(i/j).
9. [was #6]
(a) 63 8
(b) 3 2 1
(c) 2 -1 3
(d) 0 0 0
13. [was #8] The expression ++i is equivalent to (i += 1). The value of both expressions is i after the increment has been performed.
Answers to Selected Programming Projects
14、
2. [was #4]
#include
int main(void)
{
int n;
printf("Enter a three-digit number: ");
scanf("%d", &n);
printf("The reversal is: %d%d%d\n", n % 10, (n / 10) % 10, n / 100);
return 0;
}
Chapter 5
Answers to Selected Exercises
2. [was #2]
(a) 1
(b) 1
(c)
15、1
(d) 1
4. [was #4] (i > j) - (i < j)
6. [was #12] Yes, the statement is legal. When n is equal to 5, it does nothing, since 5 is not equal to –9.
10. [was #16] The output is
onetwo
since there are no break statements after the cases.
Answers to Selected Programming Projects
2. [was #6
16、]
#include
int main(void)
{
int hours, minutes;
printf("Enter a 24-hour time: ");
scanf("%d:%d", &hours, &minutes);
printf("Equivalent 12-hour time: ");
if (hours == 0)
printf("12:%.2d AM\n", minutes);
else if (hours < 12)
printf("%d:%.2d AM\n", hours,
17、 minutes);
else if (hours == 12)
printf("%d:%.2d PM\n", hours, minutes);
else
printf("%d:%.2d PM\n", hours - 12, minutes);
return 0;
}
4. [was #8; modified]
#include
int main(void)
{
int speed;
printf("Enter a wind speed in knots: ");
scanf("%d", &sp
18、eed);
if (speed < 1)
printf("Calm\n");
else if (speed <= 3)
printf("Light air\n");
else if (speed <= 27)
printf("Breeze\n");
else if (speed <= 47)
printf("Gale\n");
else if (speed <= 63)
printf("Storm\n");
else
printf("Hurricane\n");
return 0;
19、}
6. [was #10]
#include
int main(void)
{
int check_digit, d, i1, i2, i3, i4, i5, j1, j2, j3, j4, j5,
first_sum, second_sum, total;
printf("Enter the first (single) digit: ");
scanf("%1d", &d);
printf("Enter first group of five digits: ");
scanf("%1d%1d%1d%1d%
20、1d", &i1, &i2, &i3, &i4, &i5);
printf("Enter second group of five digits: ");
scanf("%1d%1d%1d%1d%1d", &j1, &j2, &j3, &j4, &j5);
printf("Enter the last (single) digit: ");
scanf("%1d", &check_digit);
first_sum = d + i2 + i4 + j1 + j3 + j5;
second_sum = i1 + i3 + i5 + j2 + j4;
21、total = 3 * first_sum + second_sum;
if (check_digit == 9 - ((total - 1) % 10))
printf("VALID\n");
else
printf("NOT VALID\n");
return 0;
}
10. [was #14]
#include
int main(void)
{
int grade;
printf("Enter numerical grade: ");
scanf("%d", &grade);
22、 if (grade < 0 || grade > 100) {
printf("Illegal grade\n");
return 0;
}
switch (grade / 10) {
case 10:
case 9: printf("Letter grade: A\n");
break;
case 8: printf("Letter grade: B\n");
break;
case 7: printf("Letter grade: C\n");
23、 break;
case 6: printf("Letter grade: D\n");
break;
case 5:
case 4:
case 3:
case 2:
case 1:
case 0: printf("Letter grade: F\n");
break;
}
return 0;
}
Chapter 6
Answers to Selected Exercises
4. [was #10] (c) is
24、not equivalent to (a) and (b), because i is incremented before the loop body is executed.
10. [was #12] Consider the following while loop:
while (…) {
…
continue;
…
}
The equivalent code using goto would have the following appearance:
while (…) {
…
goto loop_end;
…
loop_e
25、nd: ; /* null statement */
}
12. [was #14]
for (d = 2; d * d <= n; d++)
if (n % d == 0)
break;
The if statement that follows the loop will need to be modified as well:
if (d * d <= n)
printf("%d is divisible by %d\n", n, d);
else
printf("%d is prime\n", n);
14. [was #16] The
26、problem is the semicolon at the end of the first line. If we remove it, the statement is now correct:
if (n % 2 == 0)
printf("n is even\n");
Answers to Selected Programming Projects
2. [was #2]
#include
int main(void)
{
int m, n, remainder;
printf("Enter two integers:
27、");
scanf("%d%d", &m, &n);
while (n != 0) {
remainder = m % n;
m = n;
n = remainder;
}
printf("Greatest common divisor: %d\n", m);
return 0;
}
4. [was #4]
#include
int main(void)
{
float commission, value;
printf("Enter value of trade: ")
28、;
scanf("%f", &value);
while (value != 0.0f) {
if (value < 2500.00f)
commission = 30.00f + .017f * value;
else if (value < 6250.00f)
commission = 56.00f + .0066f * value;
else if (value < 20000.00f)
commission = 76.00f + .0034f * value;
else if (value
29、 < 50000.00f)
commission = 100.00f + .0022f * value;
else if (value < 500000.00f)
commission = 155.00f + .0011f * value;
else
commission = 255.00f + .0009f * value;
if (commission < 39.00f)
commission = 39.00f;
printf("Commission: $%.2f\n\n", commi
30、ssion);
printf("Enter value of trade: ");
scanf("%f", &value);
}
return 0;
}
6. [was #6]
#include
int main(void)
{
int i, n;
printf("Enter limit on maximum square: ");
scanf("%d", &n);
for (i = 2; i * i <= n; i += 2)
printf("%d\n", i * i);
31、
return 0;
}
8. [was #8]
#include
int main(void)
{
int i, n, start_day;
printf("Enter number of days in month: ");
scanf("%d", &n);
printf("Enter starting day of the week (1=Sun, 7=Sat): ");
scanf("%d", &start_day);
/* print any leading "blank dates" */
32、for (i = 1; i < start_day; i++)
printf(" ");
/* now print the calendar */
for (i = 1; i <= n; i++) {
printf("%3d", i);
if ((start_day + i - 1) % 7 == 0)
printf("\n");
}
return 0;
}
Chapter 7
Answers to Selected Exercises
3. [was #4] (b) is not legal.
4
33、. [was #6] (d) is illegal, since printf requires a string, not a character, as its first argument.
10. [was #14] unsigned int, because the (int) cast applies only to j, not j * k.
12. [was #16] The value of i is converted to float and added to f, then the result is converted to double and stored
34、 in d.
14. [was #18] No. Converting f to int will fail if the value stored in f exceeds the largest value of type int.
Answers to Selected Programming Projects
1. [was #2] short int values are usually stored in 16 bits, causing failure at 182. int and long int values are usually stored in 32 bi
35、ts, with failure occurring at 46341.
2. [was #8]
#include
int main(void)
{
int i, n;
char ch;
printf("This program prints a table of squares.\n");
printf("Enter number of entries in table: ");
scanf("%d", &n);
ch = getchar();
/* dispose of new-line characte
36、r following number of entries */
/* could simply be getchar(); */
for (i = 1; i <= n; i++) {
printf("%10d%10d\n", i, i * i);
if (i % 24 == 0) {
printf("Press Enter to continue...");
ch = getchar(); /* or simply getchar(); */
}
}
return 0;
}
5. [was
37、#10]
#include
#include
int main(void)
{
int sum = 0;
char ch;
printf("Enter a word: ");
while ((ch = getchar()) != \n)
switch (toupper(ch)) {
case D: case G:
sum += 2; break;
case B: case C: case M: case P:
sum += 3; bre
38、ak;
case F: case H: case V: case W: case Y:
sum += 4; break;
case K:
sum += 5; break;
case J: case X:
sum += 8; break;
case Q: case Z:
sum += 10; break;
default:
sum++; break;
}
printf("Scrabble value: %d\n", s
39、um);
return 0;
}
6. [was #12]
#include
int main(void)
{
printf("Size of int: %d\n", (int) sizeof(int));
printf("Size of short: %d\n", (int) sizeof(short));
printf("Size of long: %d\n", (int) sizeof(long));
printf("Size of float: %d\n", (int) sizeof(float));
prin
40、tf("Size of double: %d\n", (int) sizeof(double));
printf("Size of long double: %d\n", (int) sizeof(long double));
return 0;
}
Since the type of a sizeof expression may vary from one implementation to another, its necessary in C89 to cast sizeof expressions to a known type before printing t
41、hem. The sizes of the basic types are small numbers, so its safe to cast them to int. (In general, however, its best to cast sizeof expressions to unsigned long and print them using %lu.) In C99, we can avoid the cast by using the %zu conversion specification.
Chapter 8
Answers to Selected Exerc
42、ises
1. [was #4] The problem with sizeof(a) / sizeof(t) is that it cant easily be checked for correctness by someone reading the program. (The reader would have to locate the declaration of a and make sure that its elements have type t.)
2. [was #8] To use a digit d (in character form) as a subsc
43、ript into the array a, we would write a[d-0]. This assumes that digits have consecutive codes in the underlying character set, which is true of ASCII and other popular character sets.
7. [was #10]
const int segments[10][7] = {{1, 1, 1, 1, 1, 1},
{0, 1, 1},
44、 {1, 1, 0, 1, 1, 0, 1},
{1, 1, 1, 1, 0, 0, 1},
{0, 1, 1, 0, 0, 1, 1},
{1, 0, 1, 1, 0, 1, 1},
{1, 0, 1, 1, 1, 1, 1},
{1, 1, 1},
45、 {1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 0, 1, 1}};
Answers to Selected Programming Projects
2. [was #2]
#include
int main(void)
{
int digit_count[10] = {0};
int digit;
long n;
printf("Enter a number: ");
scanf("%
46、ld", &n);
while (n > 0) {
digit = n % 10;
digit_count[digit]++;
n /= 10;
}
printf ("Digit: ");
for (digit = 0; digit <= 9; digit++)
printf("%3d", digit);
printf("\nOccurrences:");
for (digit = 0; digit <= 9; digit++)
printf("%3d", digit_count[digit
47、]);
printf("\n");
return 0;
}
5. [was #6]
#include
#define NUM_RATES ((int) (sizeof(value) / sizeof(value[0])))
#define INITIAL_BALANCE 100.00
int main(void)
{
int i, low_rate, month, num_years, year;
double value[5];
printf("Enter interest rate: ");
scanf
48、("%d", &low_rate);
printf("Enter number of years: ");
scanf("%d", &num_years);
printf("\nYears");
for (i = 0; i < NUM_RATES; i++) {
printf("%6d%%", low_rate + i);
value[i] = INITIAL_BALANCE;
}
printf("\n");
for (year = 1; year <= num_years; year++) {
printf("%
49、3d ", year);
for (i = 0; i < NUM_RATES; i++) {
for (month = 1; month <= 12; month++)
value[i] += ((double) (low_rate + i) / 12) / 100.0 * value[i];
printf("%7.2f", value[i]);
}
printf("\n");
}
return 0;
}
8. [was #12]
#include
#define
50、 NUM_QUIZZES 5
#define NUM_STUDENTS 5
int main(void)
{
int grades[NUM_STUDENTS][NUM_QUIZZES];
int high, low, quiz, student, total;
for (student = 0; student < NUM_STUDENTS; student++) {
printf("Enter grades for student %d: ", student + 1);
for (quiz = 0; quiz < NUM_QUIZZES
51、; quiz++)
scanf("%d", &grades[student][quiz]);
}
printf("\nStudent Total Average\n");
for (student = 0; student < NUM_STUDENTS; student++) {
printf("%4d ", student + 1);
total = 0;
for (quiz = 0; quiz < NUM_QUIZZES; quiz++)
total += grades[student][quiz
52、];
printf("%3d %3d\n", total, total / NUM_QUIZZES);
}
printf("\nQuiz Average High Low\n");
for (quiz = 0; quiz < NUM_QUIZZES; quiz++) {
printf("%3d ", quiz + 1);
total = 0;
high = 0;
low = 100;
for (student = 0; student < NUM_STUDENTS; student++) {
53、
total += grades[student][quiz];
if (grades[student][quiz] > high)
high = grades[student][quiz];
if (grades[student][quiz] < low)
low = grades[student][quiz];
}
printf("%3d %3d %3d\n", total / NUM_STUDENTS, high, low);
}
return 0;
}
C
54、hapter 9
Answers to Selected Exercises
2. [was #2]
int check(int x, int y, int n)
{
return (x >= 0 && x <= n - 1 && y >= 0 && y <= n - 1);
}
4. [was #4]
int day_of_year(int month, int day, int year)
{
int num_days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int day_count
55、 = 0, i;
for (i = 1; i < month; i++)
day_count += num_days[i-1];
/* adjust for leap years, assuming they are divisible by 4 */
if (year % 4 == 0 && month > 2)
day_count++;
return day_count + day;
}
Using the expression year % 4 == 0 to test for leap years is not complet
56、ely correct. Centuries are special cases: if a year is a multiple of 100, then it must also be a multiple of 400 in order to be a leap year. The correct test is
year % 4 == 0 && (year % 100 != 0 || year % 400 == 0)
6. [was #6; modified]
int digit(int n, int k)
{
int i;
for (i = 1; i <
57、k; i++)
n /= 10;
return n % 10;
}
8. [was #8] (a) and (b) are valid prototypes. (c) is illegal, since it doesnt specify the type of the parameter. (d) incorrectly specifies that f returns an int value in C89; in C99, omitting the return type is illegal.
10. [was #10]
(a)
int largest
58、(int a[], int n)
{
int i, max = a[0];
for (i = 1; i < n; i++)
if (a[i] > max)
max = a[i];
return max;
}
(b)
int average(int a[], int n)
{
int i, avg = 0;
for (i = 0; i < n; i++)
avg += a[i];
return avg / n;
}
(c)
int num_positive(int a[], int n)
{
59、
int i, count = 0;
for (i = 0; i < n; i++)
if (a[i] > 0)
count++;
return count;
}
15. [was #12; modified]
double median(double x, double y, double z)
{
double result;
if (x <= y)
if (y <= z) result = y;
else if (x <= z) result = z;
else result = x;
60、
else {
if (z <= y) result = y;
else if (x <= z) result = x;
else result = z;
}
return result;
}
17. [was #14]
int fact(int n)
{
int i, result = 1;
for (i = 2; i <= n; i++)
result *= i;
return result;
}
19. [was #16] The following program tests the
61、pb function:
#include
void pb(int n);
int main(void)
{
int n;
printf("Enter a number: ");
scanf("%d", &n);
printf("Output of pb: ");
pb(n);
printf("\n");
return 0;
}
void pb(int n)
{
if (n != 0) {
pb(n / 2);
putchar(0 + n % 2);
}
}
62、
pb prints the binary representation of the argument n, assuming that n is greater than 0. (We also assume that digits have consecutive codes in the underlying character set.) For example:
Enter a number: 53
Output of pb: 110101
A trace of pbs execution would look like this:
pb(53) finds that
63、53 is not equal to 0, so it calls
pb(26), which finds that 26 is not equal to 0, so it calls
pb(13), which finds that 13 is not equal to 0, so it calls
pb(6), which finds that 6 is not equal to 0, so it calls
pb(3), which finds that 3 is not equal to 0, so it calls
pb(1), which finds that
64、1 is not equal to 0, so it calls
pb(0), which finds that 0 is equal to 0, so it returns, causing
pb(1) to print 1 and return, causing
pb(3) to print 1 and return, causing
pb(6) to print 0 and return, causing
pb(13) to print 1 and return, causing
pb(26) to print 0 and return, causing
pb(53)
65、to print 1 and return.
Chapter 10
Answers to Selected Exercises
1. [was #2] (a) a, b, and c are visible.
(b) a, and d are visible.
(c) a, d, and e are visible.
(d) a and f are visible.
Answers to Selected Programming Projects
3. [was #4]
#include /* C99 only */
#include
66、
#include
#define NUM_CARDS 5
#define RANK 0
#define SUIT 1
/* external variables */
int hand[NUM_CARDS][2];
/* 0 1
____ ____
0 |____|____|
1 |____|____|
2 |____|____|
3 |____|____|
4 |____|____|
rank suit
*/
bool straight, flush, four, three;
int pairs; /* can be 0, 1, or 2 */
/* protot
C語言程序設計 現(xiàn)代方法 第二版 習題答案 C Programming_ A Modern Approach
