C语言程序设计现代方法第二版习题答案CProgrammingAModernApproach

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) parkinson's law:work expands so as to fill the timeavailable for its completion.3. was #4#include <stdio.h>int main(void) int he

2、ight = 8, length = 12, 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 here&#

3、39;s one possible program:#include <stdio.h>int main(void) int i, j, k; float 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);精品. prin

4、tf("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 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

5、 chance that you'll 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, =, (, 3, *, q, -, p, *, p, ), /, 3, and ;. answers to selected programming projects4. was #8; modified#include <stdio.h>int main(voi

6、d) 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_with_tax); return 0;the amount_with_tax variable is unnecessary. if we remove it

7、, the program is slightly shorter:#include <stdio.h>精品.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 selected exercises2

8、. 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 projects1. was #4; modified #include <stdi

9、o.h>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%.2dn", year, month, day); return 0;3. was #6; modified 精品.#include <stdio.h>int main(void) in

10、t prefix, group, publisher, item, check_digit; printf("enter isbn: "); scanf("%d-%d-%d-%d-%d", &prefix, &group, &publisher, &item, &check_digit); printf("gs1 prefix: %dn", prefix); printf("group identifier: %dn", group); printf("publis

11、her 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: %dnpublisher code: %dnitem number: %dncheck digit: %dn", prefix, gro

12、up, 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 of -(i/j) is always 1, regardless of the implementation

13、. 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 0 精品.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 programmi

14、ng projects2. was #4 #include <stdio.h>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

15、 (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 onetwosince there are no break statements after the cases. answers to selected programming projects精品.2. was #6 #include <st

16、dio.h>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 amn", minutes); else if (hours < 12) printf("%d:%.2d amn",

17、hours, minutes); else if (hours = 12) printf("%d:%.2d pmn", hours, minutes); else printf("%d:%.2d pmn", hours - 12, minutes); return 0;4. was #8; modified #include <stdio.h>int main(void) int speed; printf("enter a wind speed in knots: "); scanf("%d", &a

18、mp;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");精品. else printf("hurrican

19、en"); return 0;6. was #10 #include <stdio.h>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: ")

20、; scanf("%1d%1d%1d%1d%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", &am

21、p;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("validn"); else printf("not validn"); return 0;10. was #14 #include <stdio.h>int main(void) int gr

22、ade; printf("enter numerical grade: ");精品. scanf("%d", &grade); if (grade < 0 | 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"); bre

23、ak; case 7: printf("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 exercises4. was #10 (c) is not equivalent to (a) and (b), b

24、ecause i is incremented before the loop body is executed. 10. was #12 consider the following while loop: while () continue; 精品.the equivalent code using goto would have the following appearance:while () goto loop_end; loop_end: ; /* null statement */12. was #14 for (d = 2; d * d <= n; d+) if (n %

25、 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 %dn", n, d);else printf("%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

26、now correct: if (n % 2 = 0) printf("n is evenn");answers to selected programming projects2. was #2 #include <stdio.h>int main(void) int m, n, remainder; printf("enter two integers: "); scanf("%d%d", &m, &n); while (n != 0) remainder = m % n; m = n; n = rem

27、ainder;精品. printf("greatest common divisor: %dn", m); return 0;4. was #4 #include <stdio.h>int main(void) float commission, value; printf("enter value of trade: "); scanf("%f", &value); while (value != 0.0f) if (value < 2500.00f) commission = 30.00f + .017f

28、 * 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 = 155.00f + .0011f * value; else commiss

29、ion = 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 精品.#include <stdio.h>int main(void) int i, n; printf("ent

30、er limit on maximum square: "); scanf("%d", &n); for (i = 2; i * i <= n; i += 2) printf("%dn", i * i); return 0;8. was #8 #include <stdio.h>int main(void) int i, n, start_day; printf("enter number of days in month: "); scanf("%d", &n); pr

31、intf("enter starting day of the week (1=sun, 7=sat): "); scanf("%d", &start_day); /* print any leading "blank dates" */ for (i = 1; i < start_day; i+) printf(" "); /* now print the calendar */ for (i = 1; i <= n; i+) printf("%3d", i); if (s

32、tart_day + i - 1) % 7 = 0) printf("n"); return 0;chapter 7精品.answers 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 first argument. 10. was #14 unsigned int, because the (int) cast applies only to j, not j

33、* 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 value of type int. answers to selected programming projects1. was #2 short int valu

34、es 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 <stdio.h>int main(void) int i, n; char ch; printf("this program prints a table of squares.n"); printf("enter numbe

35、r 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.&qu

36、ot;); ch = getchar(); /* or simply getchar(); */ 精品. return 0;5. was #10 #include <ctype.h>#include <stdio.h>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; br

37、eak; 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': case 'x': sum += 8; break; case 'q

38、': case 'z': sum += 10; break; default: sum+; break; printf("scrabble value: %dn", sum); return 0;6. was #12 #include <stdio.h>int main(void)精品. printf("size of int: %dn", (int) sizeof(int); printf("size of short: %dn", (int) sizeof(short); printf(&quo

39、t;size of long: %dn", (int) sizeof(long); printf("size of 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 fro

40、m one implementation to another, it's necessary in c89 to cast sizeof expressions to a known type before printing them. the sizes of the basic types are small numbers, so it's safe to cast them to int. (in general, however, it's best to cast sizeof expressions to unsigned long and print

41、them using %lu.) in c99, we can avoid the cast by using the %zu conversion specification.chapter 8answers to selected exercises1. was #4 the problem with sizeof(a) / sizeof(t) is that it can't easily be checked for correctness by someone reading the program. (the reader would have to locate the

42、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 subscript 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 char

43、acter sets. 7. was #10 const 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, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,精品. 1, 1, 1, 1, 0, 1, 1;answers to selected programming projects2. was #2 #include <stdio

44、.h>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; printf ("digit: "); for (digit = 0; digit <= 9; digit+) printf("%3d", digit); print

45、f("noccurrences:"); for (digit = 0; digit <= 9; digit+) printf("%3d", digit_countdigit); printf("n"); return 0;5. was #6 #include <stdio.h>#define num_rates (int) (sizeof(value) / sizeof(value0)#define initial_balance 100.00int main(void) int i, low_rate, month

46、, num_years, year; double value5;精品. 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 +

47、i); valuei = initial_balance; printf("n"); for (year = 1; year <= num_years; year+) 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); prin

48、tf("n"); return 0;8. was #12 #include <stdio.h>#define num_quizzes 5#define num_students 5int main(void) int gradesnum_studentsnum_quizzes; int high, low, quiz, student, total; for (student = 0; student < num_students; student+) printf("enter grades for student %d: ", st

49、udent + 1); for (quiz = 0; quiz < num_quizzes; quiz+) scanf("%d", &gradesstudentquiz); 精品. printf("nstudent total averagen"); for (student = 0; student < num_students; student+) printf("%4d ", student + 1); total = 0; for (quiz = 0; quiz < num_quizzes; quiz

50、+) 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 < num_students; student+

51、) total += gradesstudentquiz; if (gradesstudentquiz > high) high = gradesstudentquiz; if (gradesstudentquiz < low) low = gradesstudentquiz; printf("%3d %3d %3dn", total / num_students, high, low); return 0;chapter 9answers to selected exercises2. was #2 int check(int x, int y, int n)

52、 return (x >= 0 && x <= n - 1 && y >= 0 && y <= n - 1);4. was #4 int day_of_year(int month, int day, int year)精品. int num_days = 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31; int day_count = 0, i; for (i = 1; i < month; i+) day_count += num_daysi-1; /* adjust

53、 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 completely correct. centuries are special cases: if a year is a multiple of 100, then it must also be a mu

54、ltiple 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 < k; i+) n /= 10; return n % 10;8. was #8 (a) and (b) are valid prototypes. (c) is illegal, since it doesn't

55、 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 < n; i+)精品. if (ai > max) max = ai; return max;(b)int average(int a, int n) int

56、 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 < n; i+) if (ai > 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

57、; else if (x <= z) result = z; else result = x; else if (z <= y) result = y; else if (x <= z) result = x; else result = z;精品. 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 <stdio.h>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