c语言程序设计：现代方法第二版习题答案-文档

1、Chapter 2Answers to Selected Exercises2. was #2 (a) The program contains one directive (#include) and four statements (three calls of printf and one return).(b)Parkinsons Law:Work expands so as to fill the time available for its completion.3. was #4#include int main(void)int height = 8, length = 12,

2、 width = 10, volume; volume = height * length * width;printf(Dimensions: %dx%dx%dn, length, width, height); printf(Volume (cubic inches): %dn, volume);printf(Dimensional weight (pounds): %dn, (volume + 165) / 166); return 0;4. was #6 Heres one possible program:#include int main(void)int i, j, k;floa

3、t x, y, z; printf(Value of i: %dn, i); printf(Value of j: %dn, j); printf(Value of k: %dn, k); printf(Value of x: %gn, x); printf(Value of y: %gn, y); printf(Value of z: %gn, z);return 0;When compiled using GCC and then executed, this program produced the following output:Value of i: 5618848Value of

4、 j: 0Value of k: 6844404Value of x: 3.98979e-34Value of y: 9.59105e-39Value of z: 9.59105e-39The values printed depend on many factors, so the chance that youll get exactly these numbers is small.5. was #10 (a) is not legal because 100_bottles begins with a digit.8. was #12 There are 14 tokens: a, =

5、, (, 3, *, q, -, p, *, p, ), /, 3, and ;.Answers to Selected Programming Projects4. was #8; modified#include 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: $%.2fn, amount_wi

6、th_tax); return 0;The amount_with_tax variable is unnecessary. 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: $%.2fn, original_amount * 1.05f); return 0;Chapter 3Answers to Se

7、lected Exercises2. was #2(a) printf(%-8.1e, x);(b) printf(%10.6e, x);(c) printf(%-8.3f, x);(d) printf(%6.0f, x);respectively.5. was #8 The values of x, i, and y will be 12.3, 45, and .6,Answers to Selected Programming Projects1. was #4; modified#include int main(void)int month, day, year; printf(Ent

8、er a date (mm/dd/yyyy): ); scanf(%d/%d/%d, &month, &day, &year);printf(You entered the date %d%.2d%.2dn, year, month, day); return 0;3. was #6; modified#include int main(void)int prefix, group, publisher, item, check_digit; printf(Enter ISBN: );scanf(%d-%d-%d-%d-%d, &prefix, &group, &publisher, &ite

9、m, &check_digit);printf(GS1 prefix: %dn, prefix); printf(Group identifier: %dn, group); printf(Publisher code: %dn, publisher); printf(Item number: %dn, item); printf(Check digit: %dn, check_digit);/* The five printf calls can be combined as follows: printf(GS1 prefix: %dnGroup identifier: %dnPublis

10、her code: %dnItem number: %dnCheck digit: %dn, prefix, group, publisher, item, check_digit);return 0;Chapter 4Answers to Selected Exercises2. 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 o

11、f -(i/j) is always 1, regardless of the implementation.In C99, on the other hand, 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 0 013. was #8 The expression +i is equivalent to (i += 1). The value of both expressions is i after the increment has b

12、een performed.Answers to Selected Programming Projects2. was #4#include int main(void)int n;printf(Enter a three-digit number: );scanf(%d, &n);printf(The reversal is: %d%d%dn, n % 10, (n / 10) % 10, n / 100);return 0;Chapter 5Answers to Selected Exercises2. was #2(a) 1(b) 1(c) 1(d) 14. was #4 (i j)

13、- (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 isonetwo since there are no break statements after the cases.Answers to Selected Programming Projects2. was #6#include int main(void)int hours, minutes;printf(Enter

14、 a 24-hour time: ); scanf(%d:%d, &hours, &minutes); printf(Equivalent 12-hour time: );if (hours = 0)printf(12:%.2d AMn, minutes);else if (hours 12) printf(%d:%.2d AMn, hours, minutes);else if (hours = 12) printf(%d:%.2d PMn, hours, minutes);elseprintf(%d:%.2d PMn, hours - 12, minutes); return 0;4. w

15、as #8; modified#include int main(void)int speed;printf(Enter a wind speed in knots: ); scanf(%d, &speed);if (speed 1)printf(Calmn);else if (speed = 3)printf(Light airn);else if (speed = 27)printf(Breezen);else if (speed = 47)printf(Galen);else if (speed = 63)printf(Stormn);elseprintf(Hurricanen);ret

16、urn 0;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%1d, &i1, &i2, &i3, &i4, &i5); printf(Enter second gr

17、oup 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; total = 3 * first_sum + second_sum; if (check_digit = 9 - (total - 1) % 10) printf(VAL

18、IDn);elseprintf(NOT VALIDn);return 0;10. was #14#include int main(void)int grade;printf(Enter numerical grade: );scanf(%d, &grade);if (grade 100) printf(Illegal graden); return 0;switch (grade / 10) case 10:case 9: printf(Letter grade: An);break;case 8: printf(Letter grade: Bn);break;case 7: printf(

19、Letter grade: Cn);break;case 6: printf(Letter grade: Dn);break;case 5: case 4:case 3:case 2:case 1:case 0: printf(Letter grade: Fn); break;return 0;Chapter 6Answers to Selected Exercisesincremented4. was #10 (c) is not equivalent to (a) and (b), because i is before the loop body is executed.10. was

20、#12 Consider the following while loop:while ()continue;The equivalent code using goto would have the following appearance:while ( ) goto loop_end;loop_end: ; /* null statement */12. was #14for (d = 2; d * d = n; d+)if (n % d = 0)break;The if statement that follows the loop will need to be modified a

21、s well: if (d * d = n)printf(%d is divisible by %dn, n, d);elseprintf(%d is primen, n);14. was #16 The 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 evenn);Answers to Selected Programming Projects2. was #2 #include int

22、 main(void) int m, n, remainder; printf(Enter two integers: ); scanf(%d%d, &m, &n); while (n != 0) remainder = m % n; m = n;n = remainder;printf(Greatest common divisor: %dn, m); return 0;4. was #4#include int main(void) float commission, value; printf(Enter value of trade: ); scanf(%f, &value);whil

23、e (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 50000.00f) commission = 100.00f + .0022f * value;else if (value 500000.00f) commission

24、 = 155.00f + .0011f * value;elsecommission = 255.00f + .0009f * value; if (commission 39.00f)commission = 39.00f; printf(Commission: $%.2fnn, commission); 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: ); sca

25、nf(%d, &n);for (i = 2; i * i = n; i += 2) printf(%dn, i * i);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 */for (i

26、= 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 7Answers to Selected Exercises3. was #4 (b) is not legal.4. was #6 (d) is illegal, since printf requires a string, not a character, as its fir

27、st 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 in d.14. was #18 No. Converting f to int will fail if the value stored in f exceeds the largest

28、 value of type int.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 bits, 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 n

29、umber of entries in table: );scanf(%d, &n);ch = getchar();/* dispose of new-line character following number of entries */* could simply be getchar(); */for (i = 1; i = n; i+) printf(%10d%10dn, i, i * i);if (i % 24 = 0) printf(Press Enter to continue.);ch = getchar(); /* or simply getchar(); */return

30、 0;5. was #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; break;case F: case H: case V: case W: case Y:sum += 4; break;case K:sum += 5; break;case J: c

31、ase X:sum += 8; break;case Q: case Z:sum += 10; break;default:sum+; break;printf(Scrabble value: %dn, sum);return 0;6. was #12#include int main(void)printf(Size of int: %dn, (int) sizeof(int);printf(Size of short: %dn, (int) sizeof(short); printf(Size of long: %dn, (int) sizeof(long); printf(Size of

32、 float: %dn, (int) sizeof(float);printf(Size of double: %dn, (int) sizeof(double);printf(Size of long double: %dn, (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 be

33、fore printing them. 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 8Answers to Selec

34、ted Exercises1. 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 makesure that its elements have type t .)2. was #8 To use a digit d (in character form) as a subs

35、cript into the array a, we would write ad-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 #10const int segments107 = 1, 1, 1, 1, 1, 1,0, 1, 1,1, 1, 0, 1, 1, 0, 1,1, 1, 1, 1, 0, 0, 1,0, 1, 1, 0, 0, 1, 1

36、,1, 0, 1, 1, 0, 1, 1,1, 0, 1, 1, 1, 1, 1,1, 1, 1,1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1;Answers to Selected Programming Projects2. was #2#include int main(void)int digit_count10 = 0;int digit;long n;printf(Enter a number: ); scanf(%ld, &n);while (n 0) digit = n % 10; digit_countdigit+;n /= 10;prin

37、tf (Digit: );for (digit = 0; digit = 9; digit+) printf(%3d, digit);printf(nOccurrences:);for (digit = 0; digit = 9; digit+) printf(%3d, digit_countdigit);printf(n);return 0;5. was #6#include #define NUM_RATES (int) (sizeof(value) / sizeof(value0) #define INITIAL_BALANCE 100.00int main(void)int i, lo

38、w_rate, month, num_years, year; double value5;printf(Enter interest rate: ); scanf(%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);valuei = INITIAL_BALANCE; printf(n);for (year = 1; year = num_years; year+)

39、 printf(%3d , year);for (i = 0; i NUM_RATES; i+) for (month = 1; month = 12; month+)valuei += (double) (low_rate + i) / 12) / 100.0 * valuei; printf(%7.2f, valuei);printf(n);return 0;8. was #12#include #define NUM_QUIZZES 5#define NUM_STUDENTS 5int main(void)int gradesNUM_STUDENTSNUM_QUIZZES;int hig

40、h, 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; quiz+) scanf(%d, &gradesstudentquiz);printf(nStudent Total Averagen);for (student = 0; student NUM_STUDENTS; student+) printf(%4d , stude

41、nt + 1);total = 0;for (quiz = 0; quiz NUM_QUIZZES; quiz+)total += gradesstudentquiz;printf(%3d %3dn, total, total / NUM_QUIZZES); printf(nQuiz Average High Lown);for (quiz = 0; quiz NUM_QUIZZES; quiz+) printf(%3d , quiz + 1);total = 0;high = 0;low = 100;for (student = 0; student high)high = gradesst

42、udentquiz;if (gradesstudentquiz = 0 & x = 0 & y = n - 1);4. was #4int 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 = 0, i;for (i = 1; i 2)day_count+;return day_count + day;Using the expression year %4 = 0 to test for leap years

43、is not completely 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 isyear % 4 = 0 & (year % 100 != 0 | year % 400 = 0)6. was #6; modifiedint digit(int n, int k)int i;for (i = 1; i k; i+)n /= 10;r

44、eturn n % 10;8. was #8 (a) and (b) are valid prototypes. (c) is illegal, since itdoesnt 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(int a, int n)int i, max = a0;for (i = 1; i ma

45、x) max = ai;return max;(b)int average(int a, int n) int i, avg = 0;for (i = 0; i n; i+) avg += ai;return avg / n;(c)int num_positive(int a, int n) int i, count = 0;for (i = 0; i 0)count+;return count;15. was #12; modifieddouble median(double x, double y, double z) double result;if (x = y)if (y = z)

46、result = y;else if (x = z) result = z;else result = x;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 pb function: #include vo

47、id 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);pb prints the binary representation of the argument n, assuming that n is greater than 0.(We also assume that digits h

48、ave consecutivecodes inthe underlying character set.) For example:Enter a number: 53 Output of pb: 110101A trace of pbs execution would look like this:pb(53) finds that 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 equ

49、al 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 callspb(1), which finds that 1 is not equal to 0, so it callspb(0), which finds that 0is equal to 0, soit returns, causingpb(1) to print 1 and return, causing pb(3) to pri

50、nt 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, causingpb(53) to print 1 and return.Chapter 10Answers to Selected Exercises1. was #2 (a) a, b, and c are visible.(b) a, and d are visible.(c) a, d, and e are visible.(d)

51、a and f are visible.Answers to Selected Programming Projects3. was #4#include /* C99 only */#include #include #define NUM_CARDS 5#define RANK 0#define SUIT 1/* external variables */int handNUM_CARDS2;/* 0 10 |1 |2 |3 |4 |rank suit bool straight, flush, four, three; int pairs; /* can be 0, 1, or 2 */

52、 /* prototypes */ void read_cards(void); void analyze_hand(void); void print_result(void);* main: Calls read_cards, analyze_hand, and print_result * repeatedly. * int main(void)for (;) read_cards(); analyze_hand(); print_result();* read_cards: Reads the cards into the external variable * * hand; checks for bad cards and duplicate * * cards. * void read_cards(void)c