c语言程序设计现代方法第二版习题答案.docx

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 int main(void) int heigh

2、t = 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 Heres one possible program:#include int main(

3、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(Value of z: %gn, z); return 0;When compiled using GCC and then executed, this program produced the following output:Val

4、ue 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 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 T

5、here are 14 tokens: a, =, (, 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

6、tax added: $%.2fn, amount_with_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);

7、return 0;Chapter 3Answers to Selected Exercises2. 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 int main

8、(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 #include int main(void) int prefix, group, publisher, item, check_digit; printf(Enter ISBN: ); scanf(%d-%d-

9、%d-%d-%d, &prefix, &group, &publisher, &item, &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: prin

10、tf(GS1 prefix: %dnGroup identifier: %dnPublisher 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 t

11、he implementation. On the other hand, the value of -(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 0 13. was #8 The expression +i is equivalent to (i += 1). The

12、value of both expressions is i after the increment has been 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

13、Selected Exercises2. was #2 (a) 1 (b) 1 (c) 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 onetwosince there are no break statements after the cases. Answers to Selected Programming P

14、rojects2. was #6 #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 AMn, minutes); else if (hours 12) printf(%d:%.2d AMn, hours, minutes); else if (hours = 12) printf(%d:%.2d PMn

15、, hours, minutes); else printf(%d:%.2d PMn, hours - 12, minutes); return 0;4. was #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 i

16、f (speed = 47) printf(Galen); else if (speed = 63) printf(Stormn); else printf(Hurricanen); return 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 fir

17、st group of five digits: ); 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, &check_digit); first_sum = d + i2 + i4 + j1 + j3 + j5; second_sum = i1 + i3 +

18、 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 int main(void) int grade; printf(Enter numerical grade: ); scanf(%d, &grade); if (grade 100) printf(Illegal graden); return 0; switch (gra

19、de / 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; case 5: case 4: case 3: case 2: case 1: case 0: printf(Letter grade: Fn); break; return 0;Chapter 6Answers to Selected

20、 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 () continue; The equivalent code using goto would have the following appearance:while () goto loop_end; loop_end: ; /* null state

21、ment */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 %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 rem

22、ove it, the statement is now correct: if (n % 2 = 0) printf(n is evenn);Answers to Selected Programming Projects2. was #2 #include 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 div

23、isor: %dn, m); return 0;4. was #4 #include 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 * value; else if (value 6250.00f) commission = 56.00f + .0066f * value; else if (value 20000.00f

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

25、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(%dn, i * i); return 0;8. was #8 #include int main(void) int i, n, start_day; printf(Enter number of da

26、ys 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 = 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); re

27、turn 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 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 a

28、nd 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 values are usually stored in 16 bits, causing failure at 18

29、2. 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 number of entries in table: ); scanf(%d, &n); ch = getchar(); /* dispose of new-line character following number