欢迎来到三一办公! | 帮助中心 三一办公31ppt.com(应用文档模板下载平台)
三一办公
全部分类
  • 办公文档>
  • PPT模板>
  • 建筑/施工/环境>
  • 毕业设计>
  • 工程图纸>
  • 教育教学>
  • 素材源码>
  • 生活休闲>
  • 临时分类>
  • ImageVerifierCode 换一换
    首页 三一办公 > 资源分类 > DOC文档下载  

    c语言程序设计现代方法(第二版)习题答案.doc

    • 资源ID:2384671       资源大小:191KB        全文页数:72页
    • 资源格式: DOC        下载积分:8金币
    快捷下载 游客一键下载
    会员登录下载
    三方登录下载: 微信开放平台登录 QQ登录  
    下载资源需要8金币
    邮箱/手机:
    温馨提示:
    用户名和密码都是您填写的邮箱或者手机号,方便查询和重复下载(系统自动生成)
    支付方式: 支付宝    微信支付   
    验证码:   换一换

    加入VIP免费专享
     
    账号:
    密码:
    验证码:   换一换
      忘记密码?
        
    友情提示
    2、PDF文件下载后,可能会被浏览器默认打开,此种情况可以点击浏览器菜单,保存网页到桌面,就可以正常下载了。
    3、本站不支持迅雷下载,请使用电脑自带的IE浏览器,或者360浏览器、谷歌浏览器下载即可。
    4、本站资源下载后的文档和图纸-无水印,预览文档经过压缩,下载后原文更清晰。
    5、试题试卷类文档,如果标题没有明确说明有答案则都视为没有答案,请知晓。

    c语言程序设计现代方法(第二版)习题答案.doc

    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 main(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 #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("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 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, =, (, 3, *, q, -, p, *, p, ), /, 3, and ;.Answers to Selected Programming Projects4. was #8; modified#include <stdio.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: $%.2fn", amount_with_tax); return 0;The amount_with_tax variable is unnecessary. If we remove it, the program 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) 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 <stdio.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#include <stdio.h> int main(void)int 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("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: %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 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 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 been performed. Answers to Selected Programming 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(d) 14. 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 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("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", 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 <= 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_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 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;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 < 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;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 () 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 as 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 <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 = remainder; 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 * 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; 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 <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#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 (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 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 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 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 number 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 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; 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; printf("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 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 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.Chapter 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 character 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, 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", &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"); 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;

    注意事项

    本文(c语言程序设计现代方法(第二版)习题答案.doc)为本站会员(laozhun)主动上传,三一办公仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对上载内容本身不做任何修改或编辑。 若此文所含内容侵犯了您的版权或隐私,请立即通知三一办公(点击联系客服),我们立即给予删除!

    温馨提示:如果因为网速或其他原因下载失败请重新下载,重复下载不扣分。




    备案号:宁ICP备20000045号-2

    经营许可证:宁B2-20210002

    宁公网安备 64010402000987号

    三一办公
    收起
    展开