c语言程序设计现代方法(第二版)习题答案.doc

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1、c语言程序设计现代方法(第二版)习题答案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 timeavailable for its completion.3. was #4#include <stdio.h> int mai

2、n(void)int height = 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

3、 #6 Heres 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);printf(

4、"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 ch

5、ance 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, =, (, 3, *, q, -, p, *, p, ), /, 3, and ;.Answers to Selected Programming Projects4. was #8; modified#include <stdio.h> int main(void)float

6、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, the progra

7、m is slightly shorter:#include <stdio.h> 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. was #2(a) print

8、f("%-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 <stdio.h> int main(void)

9、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#include <stdio.h> int main(void)int prefix, group, publisher, i

10、tem, 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("Publisher code: %dn", publisher);

11、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, group, publisher, item, check_digit);*

12、/ 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. In C99, on the other hand, the va

13、lue 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 been performed. Answers to Selected Programming Projects2. was #4#include <stdio.h>

14、 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) - (i < j)6. was #12 Y

15、es, 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 isonetwosince there are no break statements after the cases.Answers to Selected Programming Projects2. was #6#include <stdio.h> int main(void)int hours, minutes; printf("Ent

16、er 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&quo

17、t;, hours, minutes); elseprintf("%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", &speed); if (speed < 1)printf("Calmn");else if (speed

18、<= 3)printf("Light airn");else if (speed <= 27)printf("Breezen");else if (speed <= 47)printf("Galen");else if (speed <= 63)printf("Stormn");elseprintf("Hurricanen"); return 0;6. was #10#include <stdio.h> int main(void)int check_digi

19、t, 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); prin

20、tf("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 + j

21、4;total = 3 * first_sum + second_sum; if (check_digit = 9 - (total - 1) % 10)printf("VALIDn");elseprintf("NOT VALIDn"); return 0;10. was #14#include <stdio.h> int main(void)int grade; printf("Enter numerical grade: ");scanf("%d", &grade); if (grade &

22、lt; 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");break;case 7: printf("Letter grade: Cn");break;case 6: printf("Letter grade: Dn");break

23、;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), because i is incremented before the loop body is executed.10. was #12 Consider the following while loop:while () co

24、ntinue;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 as well: if (d * d <= n)printf("%d is

25、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 <stdio.

26、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 = remainder; printf("Greatest common divisor: %dn", m);return 0;4. was #4#include <stdio.h> int main(void)float commissi

27、on, value; printf("Enter value of trade: ");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 *

28、value; else if (value < 50000.00f)commission = 100.00f + .0022f * value; else if (value < 500000.00f)commission = 155.00f + .0011f * value; elsecommission = 255.00f + .0009f * value;if (commission < 39.00f)commission = 39.00f; printf("Commission: $%.2fnn", commission);printf("

29、;Enter value of trade: ");scanf("%f", &value); return 0;6. was #6#include <stdio.h> int main(void)int i, n; printf("Enter limit on maximum square: ");scanf("%d", &n); for (i = 2; i * i <= n; i += 2)printf("%dn", i * i); return 0;8. was #8

30、#include <stdio.h> 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

31、 (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 7Answers to Selected Exercises3. was #4 (b) is not legal.4. was #6 (d) is illegal, since prin

32、tf 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 in d.14. was #18 No. Converting f to int will

33、fail if the value stored in f exceeds the largest value of type int.Answers to Selected Programming Projects1. 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 <s

34、tdio.h> 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 character following number of entries */ /* could simply be getchar(); */

35、 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 0;5. was #10#include <ctype.h>#include <stdio.h> int main(void)int sum = 0;char ch; printf("Enter a word

36、: "); 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: case X:sum += 8; break;case Q: case Z:sum += 10; break;default:sum+; break; p

37、rintf("Scrabble value: %dn", sum); return 0;6. was #12#include <stdio.h> 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 floa

38、t: %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 expres

39、sions to a known type before 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.C

40、hapter 8Answers to Selected 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 make sure that its elements have type t.)2. was #8 To use a digit d (in ch

41、aracter form) as a subscript into the array a, we would write ad-0. This assumes that digits haveconsecutive 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,

42、 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, 1,1, 1, 1, 1, 0, 1, 1;Answers to Selected Programming Projects2. was #2#include <stdio.h> int main(void)int digit_count10 = 0;int digit;long n; printf("Enter a number: ");scanf("%ld", &

43、amp;n); while (n > 0) digit = n % 10;digit_countdigit+;n /= 10; printf ("Digit: ");for (digit = 0; digit <= 9; digit+)printf("%3d", digit);printf("nOccurrences:");for (digit = 0; digit <= 9; digit+)printf("%3d", digit_countdigit);printf("n")

44、; return 0;5. was #6#include <stdio.h> #define NUM_RATES (int) (sizeof(value) / sizeof(value0) #define INITIAL_BALANCE 100.00 int main(void)int i, low_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+) printf("%3d ", year);for (i = 0;