免费预览已结束,剩余1页可下载查看
下载本文档
版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报或认领
文档简介
Chapter 2 Elementary Programming1.Valid identifiers: applet, Applet, $4, apps, x, y, radiusInvalid identifiers: a+, -a, 4#R, #44, class, public, intKeywords: class, public, int 2. double miles = 100;final double KILOMETERS_PER_ MILE = 1.609;double kilometers = KILOMETERS_PER_ MILE * miles;System.out.println(kilometers);The value of kilometers is 160.9.3.There are three benefits of using constants: (1) you dont have to repeatedly type the same value; (2) the value can be changed in a single location, if necessary; (3) the program is easy to read.final int SIZE = 20;4.a = 46 / 9; = a = 5a = 46 % 9 + 4 * 4 - 2; = a = 1 + 16 2 = 15a = 45 + 43 % 5 * (23 * 3 % 2); = a = 45 + 3 * (1) = 48a %= 3 / a + 3; = a %= 3 + 3; a % = 6 = a = a % 6 = 1;d = 4 + d * d + 4; = 4 + 1.0 + 4 = 9.0d += 1.5 * 3 + (+a); = d += 4.5 + 2; d += 6.5; = d = 7.5d -= 1.5 * 3 + a+; = d -= 4.5 + 1; = d = 1 5.5 = -4.55.22-4-4016. (2 + 100) % 7 = 4. So it is Thursday.7.For byte, from -128 to 127, inclusive.For short, from -32768 to 32767, inclusive.For int, from -2147483648 to 2147483647, inclusive.For long, from -9223372036854775808 to 9223372036854775807.For float, the smallest positive float is 1.40129846432481707e-45 and the largest float is 3.40282346638528860e+38.For double, the smallest positive double is 4.94065645841246544e-324 and the largest double is 1.79769313486231570e+308d.8.25 / 4 is 6. If you want the quotient to be a floating-point number, rewrite it as 25.0 / 4.0, 25.0 / 4, or 25 / 4.0.9.Yes, the statements are correct. The printout is 25 / 4 is 625 / 4.0 is 6.253 * 2 / 4 is 13.0 * 2 / 4 is 1.510. a. 4.0 / (3.0 * (r + 34) 9 * (a + b * c) + (3.0 + d * (2 + a) / (a + b * d)11. 1.0 * m * (r * r) 12. b and c are true.13. All.14. Line 2: Missing static for the main method. Line 2: string should be String. Line 3: i is defined but not initialized before it is used in Line 5. Line 4: k is an int, cannot assign a double value to k. Lines 7-8: The string cannot be broken into two lines.15. long totalMills = System.currentTimeMillis() returns the milliseconds since Jan 1, 1970. long totalSeconds = totalMills / 1000 returns the total seconds. long totalMinutes = totalSeconds / 60 returns the total minutes. totalMinutes % 60 returns the current minute.16.Yes. Different types of numeric values can be used in the same computation through numeric conversions referred to as casting. 17. The fractional part is truncated. Casting does not change the variable being cast. 18. f is 12.5i is 1219. System.out.println(int)1); System.out.println(int)A); System.out.println(int)B); System.out.println(int)a); System.out.println(int)b); System.out.println(char)40); System.out.println(char)59); System.out.println(char)79); System.out.println(char)85); System.out.println(char)90); System.out.println(char)0X40); System.out.println(char)0X5A); System.out.println(char)0X71); System.out.println(char)0X72); System.out.println(char)0X7A);20.u345dE is wrong. It must have exactly four hex numbers.21. and ”22.i becomes 49, since the ASCII code of 1 is 49;j become 99 since (int)1 is 49 and (int)2 is 50;k becomes 97 since the ASCII code of a is 97;c becomes character z since (int) z is 90;23.char c = A;i = (int)c; / i becomes 65float f = 1000.34f;int i = (int)f; / i becomes 1000double d = 1000.34;int i = (int)d; / i becomes 1000int i = 97;char c = (char)i; / c becomes a24. bc-225. System.out.println(1 + 1); = 11System.out.println(1 + 1); = 50 (since the Unicode for 1 is 49System.out.println(1 + 1 + 1); = 111System.out.println(1 + (1 + 1); = 12System.out.println(1 + 1 + 1); = 51 26.1 + Welcome + 1 + 1 is 1Welcome 11.1 + Welcome + (1 + 1) is 1Welcome 2.1 + Welcome + (u0001 + 1) is 1Welcome 21 + Welcome + a + 1 is 1Welcome a127.Class names: Capitalize the first letter in each name.Variables and method names: Lowercase the first word, capitalize the first letter in all subsequent words.Constants: Capitalize all letters.28.public class Test /* Main method */ public static void main(String args) / Print a line System.out.println(2 % 3 = + 2 % 3); 29.Compilation errors are detected by compilers. Runtime errors occur during execution of the program. Logic errors results in incorrect results.30. The Math class is in the java.lang package. Any class
温馨提示
- 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
- 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
- 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
- 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
- 5. 人人文库网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,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
- 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
- 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。
最新文档
- 成人高等教育中的增收与收入再分配影响研究
- 教育家精神对教师职业幸福感的提升路径探讨
- 起重设备安装期间的设备检测与验收方案
- 市政建筑工程绿色节能材料的应用研究
- 2025年蚌埠高新投资集团有限公司职业经理人招聘1名备考题库及答案详解(考点梳理)
- 2025中国检验认证集团云南有限公司财务中心招聘2人备考题库含答案详解(综合卷)
- 燃气工程质量验收标准方案
- 2025新疆铁门关市优牧草业有限公司招聘23人备考题库含答案详解(黄金题型)
- 项目变更管理与调整方案
- 2026中国储备粮管理集团有限公司广西分公司招聘45人备考题库含答案详解(新)
- 【MOOC】理解马克思-南京大学 中国大学慕课MOOC答案
- 南京理工大学紫金学院《机械设计基础》2022-2023学年第一学期期末试卷
- DB13∕T 2783-2018 冬小麦小定额灌溉技术规程
- 未被列入违法失信名单承诺书
- T-CACM 1184-2019 中医内科临床诊疗指南 酒精性肝病
- 为成果而管理
- 【中考冲刺】2023年广东省梅州市中考物理模拟试卷(附答案)
- 植物的矿质营养植物生理学
- GB/T 39529-2020系统门窗通用技术条件
- 北京理工大学英语统考B试卷+答案
- DBJ51T 196-2022 四川省智慧工地建设技术标准
评论
0/150
提交评论