欢迎来到麦多课文档分享! | 帮助中心 海量文档,免费浏览,给你所需,享你所想!
麦多课文档分享
全部分类
  • 标准规范>
  • 教学课件>
  • 考试资料>
  • 办公文档>
  • 学术论文>
  • 行业资料>
  • 易语言源码>
  • ImageVerifierCode 换一换
    首页 麦多课文档分享 > 资源分类 > PPT文档下载
    分享到微信 分享到微博 分享到QQ空间

    Chapter 2 - Control Structures.ppt

    • 资源ID:379639       资源大小:202KB        全文页数:39页
    • 资源格式: PPT        下载积分:2000积分
    快捷下载 游客一键下载
    账号登录下载
    微信登录下载
    二维码
    微信扫一扫登录
    下载资源需要2000积分(如需开发票,请勿充值!)
    邮箱/手机:
    温馨提示:
    如需开发票,请勿充值!快捷下载时,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)。
    如需开发票,请勿充值!如填写123,账号就是123,密码也是123。
    支付方式: 支付宝扫码支付    微信扫码支付   
    验证码:   换一换

    加入VIP,交流精品资源
     
    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。

    Chapter 2 - Control Structures.ppt

    1、1,Chapter 2 - Control Structures,Outline 2.1 Introduction 2.2 Algorithms 2.3 Pseudocode 2.4 Control Structures 2.5 if Selection Structure 2.6 if/else Selection Structure 2.7 while Repetition Structure 2.8 Formulating Algorithms: Case Study 1: Counter-Controlled Repetition) 2.9 Formulating Algorithms

    2、 with Top-Down, Stepwise Refinement: Case Study 2: Sentinel-Controlled Repetition) 2.10 Formulating Algorithms with Top-Down, Stepwise Refinement: Case Study 3: Nested Control Structures),2,2.1 Introduction,Before writing a program Have a thorough understanding of problem Carefully plan your approac

    3、h for solving it While writing a program Know what “building blocks” are available Use good programming principles,3,2.2 Algorithms,Computing problems Solved by executing a series of actions in a specific order Algorithm a procedure determining Actions to be executed Order to be executed Example: re

    4、cipe Program control Specifies the order in which statements are executed,4,2.3 Pseudocode,Pseudocode Artificial, informal language used to develop algorithms Similar to everyday English Not executed on computers Used to think out program before coding Easy to convert into C+ program Only executable

    5、 statements No need to declare variables,5,2.4 Control Structures,Sequential execution: statements executed in order cout “Welcome “; cout “to “; cout “C+“; Transfer of control: Next statement executed not next one in sequence cout “Welcome“; cout “ to“; if (course=105)cout “ C!“; else if (course=18

    6、1)cout “ C+! “; else / Print error message and exitcerr “Unknown course“;exit(-1);,6,2.4 Control Structures,3 control structures (Bohm and Jacopini) are sufficient to express any algorithm Sequence structure Programs executed sequentially by default Selection structures if, if/else, switch Repetitio

    7、n structures while, do/while, for,gotos are bad, mmm kay?,“Go To Statement Considered Harmful“ Edsger W. Dijkstra, CACM Mar 1968 http:/www.acm.org/classics/oct95/ Communications of the ACM, Vol. 11, No. 3, March 1968, pp. 147-148,7,Example of “goto“ statement,cout “Welcome“; cout “ to“; if (course =

    8、 181) goto greet181; if (course = 105) goto greet105; cerr “nError: Unknown Course“ endl; return -1; / exit with error status greet181: cout “ C+!“ endl; goto endProgram; greet105: cout “ C!“ endl; goto endProgram; endProgram: cout “Goodbye for now“ endl; return 0; / exit successfully.,cout “Welcome

    9、“; cout “ to“; if (course = 181) cout “ C+!“ endl; else if (course = 105) cout “ C!“ endl; else cerr “nError: Unknown Course“ endl; return -1; / exit with error status cout “Goodbye for now“ endl; return 0; / exit successfully.,cout “Welcome“; cout “ to“; if (course = 181) goto greet181; if (course

    10、= 105) goto greet105; cerr “nError: Unknown Course“ endl; return -1; / exit with error status greet181: cout “ C+!“ endl; goto endProgram; greet105: cout “ C!“ endl; goto endProgram; endProgram: cout “Goodbye for now“ endl; return 0; / exit successfully.,8,Dijkstra Letter Quotes,.the quality of prog

    11、rammers is a decreasing function of the density of go to statements in the programs they produce. .the go to statement should be abolished from all “higher level“ programming languages (i.e. everything except, perhaps, plain machine code). we should . shorten the conceptual gap between the static pr

    12、ogram and the dynamic process, to make the correspondence between the program (spread out in text space) and the process (spread out in time) as trivial as possible. The go to statement as it stands is just too primitive; it is too much an invitation to make a mess of ones program,Go To Statement Co

    13、nsidered Harmful, Communications of the ACM, Vol. 11, No. 3, March 1968, pp. 147-148,Edsger W. Dijkstra 1930-2002,9,More Dijkstra Quotes,“Computer Science is no more about computers than astronomy is about telescopes.“ “A Programming Language is a tool that has profound influence on our thinking hab

    14、its.“ “The competent programmer is fully aware of the strictly limited size of his own skull; therefore he approaches the programming task in full humility, and among other things he avoids clever tricks like the plague.“ “Progress is possible only if we train ourselves to think about programs witho

    15、ut thinking of them as pieces of executable code.“ “Program testing can best show the presence of errors but never their absence.“,10,2.4 C+ keywords,Cannot be used as identifiers or variable names,Keywords common to the C and C+ programming languages,C+ only keywords,11,2.4 Control Structures,Flowc

    16、hart Graphical representation of an algorithm Special-purpose symbols connected by arrows (flowlines) Rectangle symbol (action symbol) Any type of action Oval (or circle) symbol Beginning or end (of a program, or a section of code) Single-entry/single-exit control structures (“no gotos“) Connect exi

    17、t point of one to entry point of the next Control structure stacking,12,2.5 if Selection Structure,Selection structure: Choose among alternative courses of action If students grade is greater than or equal to 60Print “Passed” if ( grade = 60 ) cout “Passed“; Indenting makes programs easier to read C

    18、+ ignores whitespace characters (tabs, spaces, etc.) Diamond symbol (decision symbol) Indicates decision is to be made Contains an expression that can be true or false Note: in C/C+, every expression has true/false value: zero implies false, nonzero implies true. Example: if ( 3 - 4) is interpreted

    19、as “true“. if structure has single-entry/single-exit,13,2.6 if/else Selection Structure,if Performs action if condition true if/else Different actions if conditions true or falseif students grade is greater than or equal to 60 print “Passed” elseprint “Failed” if ( grade = 60 ) cout “Passed“; else c

    20、out “Failed“;,14,2.6 Ternary conditional operator (?:),Three arguments: (condition, value if true, value if false)cout = 60 ? “Passed” : “Failed” );,15,Comparison,if/else if ( grade = 60 ) cout = 60 ? “Passed” : “Failed” );,16,2.6 if/else Selection Structure,Nested if/else structures One inside anot

    21、her, test for multiple cases Once condition met, other statements skipped if students grade is greater than or equal to 90 Print “A” else if students grade is greater than or equal to 80 Print “B” else if students grade is greater than or equal to 70 Print “C” else if students grade is greater than

    22、or equal to 60 Print “D” elsePrint “F”,17,2.6 if/else Selection Structure,Example if ( grade = 90 ) / 90 and above cout = 80 ) / 80-89 cout = 70 ) / 70-79 cout = 60 ) / 60-69 cout “D“; else / less than 60 cout “F“;,18,2.6 if/else Selection Structure,Compound statement Set of statements within a pair

    23、 of bracesif ( grade = 60 ) cout “Passed.n“; else cout “Failed.n“; cout “You must take this course again.n“; Without braces, cout “You must take this course again.n“; always executed Block Set of statements within braces,19,2.6 Indentation Rules,Deitel/Deitel prefer: if ( grade = 60 ) cout = 60 ) co

    24、ut “Passed.n“; else cout “Failed.n“; cout “You must take this course again.n“; Whatever you do, be consistent. (choose one way or other way, otherwise may lose points.),20,2.7 while Repetition Structure,Repetition structure Action repeated while some condition remains true while there are more items

    25、 on my shopping listPurchase next item and cross it off my list while loop repeated until condition becomes false Example int product = 2; while ( product = 1000 )product = 2 * product;,21,2.8 Formulating Algorithms (Counter-Controlled Repetition),Counter-controlled repetition Loop repeated until co

    26、unter reaches certain value Definite repetition Number of repetitions known ExampleA class of ten students took a quiz. The grades (integers in the range 0 to 100) for this quiz are available. Determine the class average on the quiz. We know there are exactly 10 scores. So, we are going to need to r

    27、epeat “something“ exactly 10 times. There may be some stuff to do first to set things up, then some stuff to do at the end to finish up, as well.,22,2.8 Pseudocode for finding class average (counter controlled repetition),ExampleA class of ten students took a quiz. The grades (integers in the range

    28、0 to 100) for this quiz are available. Determine the class average on the quiz. Pseudocode for example: Set total to zero Set grade counter to one While grade counter is less than or equal to ten Input the next grade Add the grade into the total Add one to the grade counter Set the class average to

    29、the total divided by ten Print the class average Next: C+ code for this example,23,fig02_07.cpp (1 of 2),1 / Fig. 2.7: fig02_07.cpp 2 / Class average program with counter-controlled repetition. 3 #include 4 5 using std:cout; 6 using std:cin; 7 using std:endl; 8 9 / function main begins program execu

    30、tion 10 int main() 11 12 int total; / sum of grades input by user 13 int gradeCounter; / number of grade to be entered next 14 int grade; / grade value 15 int average; / average of grades 16 17 / initialization phase 18 total = 0; / initialize total 19 gradeCounter = 1; / initialize loop counter 20,

    31、24,fig02_07.cpp (2 of 2) fig02_07.cpp output (1 of 1),21 / processing phase 22 while ( gradeCounter grade; / read grade from user 25 total = total + grade; / add grade to total 26 gradeCounter = gradeCounter + 1; / increment counter 27 28 29 / termination phase 30 average = total / 10; / integer div

    32、ision 31 32 / display result 33 cout “Class average is “ average endl; 34 35 return 0; / indicate program ended successfully 36 37 / end function main,Enter grade: 98 Enter grade: 76 Enter grade: 71 Enter grade: 87 Enter grade: 83 Enter grade: 90 Enter grade: 57 Enter grade: 79 Enter grade: 82 Enter

    33、 grade: 94 Class average is 81,25,2.9 Formulating Algorithms (Sentinel-Controlled Repetition),Suppose problem becomes: Develop a class-averaging program that will process an arbitrary number of grades each time the program is run Unknown number of students How will program know when to end? Sentinel

    34、 value Indicates “end of data entry” Loop ends when sentinel input Sentinel chosen so it cannot be confused with regular input -1 in this case,26,2.9 Formulating Algorithms (Sentinel-Controlled Repetition),Top-down, stepwise refinement Begin with pseudocode representation of “top“ (whole problem) De

    35、termine the class average for the quiz Then, divide top into smaller tasks, list in order,27,2.9 Many programs have three phases,Helps break up programs for top-down refinement Initialization Initializes the program variables Processing Input data, adjusts program variables Termination Calculate and

    36、 print the final results,28,2.9 Refinement of the first two phases.,Refine the initialization phase Initialize variablesgoes to Initialize total to zero Initialize counter to zero ProcessingInput, sum and count the quiz grades goes to Input the first grade (possibly the sentinel) While the user has

    37、not as yet entered the sentinelAdd this grade into the running totalAdd one to the grade counterInput the next grade (possibly the sentinel),29,2.9 Refinement of the termination phase,Termination Calculate and print the class average goes to If the counter is not equal to zeroSet the average to the

    38、total divided by the counterPrint the average ElsePrint “No grades were entered” Next: C+ program,30,fig02_09.cpp (1 of 3),1 / Fig. 2.9: fig02_09.cpp 2 / Class average program with sentinel-controlled repetition. 3 #include 4 5 using std:cout; 6 using std:cin; 7 using std:endl; 8 using std:fixed; 9

    39、10 #include / parameterized stream manipulators 11 12 using std:setprecision; / sets numeric output precision 13 14 / function main begins program execution 15 int main() 16 17 int total; / sum of grades 18 int gradeCounter; / number of grades entered 19 int grade; / grade value 20 21 double average

    40、; / number with decimal point for average 22 23 / initialization phase 24 total = 0; / initialize total 25 gradeCounter = 0; / initialize loop counter,31,fig02_09.cpp (2 of 3),26 27 / processing phase 28 / get first grade from user 29 cout grade; / read grade from user 31 32 / loop until sentinel va

    41、lue read from user 33 while ( grade != -1 ) 34 total = total + grade; / add grade to total 35 gradeCounter = gradeCounter + 1; / increment counter 36 37 cout grade; / read next grade 39 40 / end while 41 42 / termination phase 43 / if user entered at least one grade . 44 if ( gradeCounter != 0 ) 45

    42、46 / calculate average of all grades entered 47 average = static_cast( total ) / gradeCounter; 48,32,fig02_09.cpp (3 of 3) fig02_09.cpp output (1 of 1),49 / display average with two digits of precision 50 cout “Class average is “ setprecision( 2 ) 51 fixed average endl; 52 53 / end if part of if/els

    43、e 54 55 else / if no grades were entered, output appropriate message 56 cout “No grades were entered“ endl; 57 58 return 0; / indicate program ended successfully 59 60 / end function main,Enter grade, -1 to end: 75 Enter grade, -1 to end: 94 Enter grade, -1 to end: 97 Enter grade, -1 to end: 88 Ente

    44、r grade, -1 to end: 70 Enter grade, -1 to end: 64 Enter grade, -1 to end: 83 Enter grade, -1 to end: 89 Enter grade, -1 to end: -1 Class average is 82.50,33,2.10 Nested Control Structures,Problem statementA college has a list of test results (1 = pass, 2 = fail) for 10 students. Write a program that

    45、 analyzes the results. If more than 8 students pass, print “Raise Tuition“. Notice that Program processes 10 results Fixed number, use counter-controlled loop Two counters can be used One counts number that passed Another counts number that fail Each test result is 1 or 2 If not 1, assume 2,Conrad C

    46、omments: Probably better to do error checking on input!,34,2.10 Nested Control Structures,Top level outline Analyze exam results and decide if tuition should be raised First refinementInitialize variablesInput the ten quiz grades and count passes and failuresPrint a summary of the exam results and d

    47、ecide if tuition should be raised Refine Initialize variables to Initialize passes to zero Initialize failures to zero Initialize student counter to one,35,2.10 Nested Control Structures,Refine Input the ten quiz grades and count passes and failuresto While student counter is less than or equal to t

    48、en Input the next exam resultIf the student passedAdd one to passes Else Add one to failuresAdd one to student counter ,36,2.10 Nested Control Structures,Refine Print a summary of the exam results and decide if tuition should be raised to Print the number of passes Print the number of failures If more than eight students passed Print “Raise tuition” Next: C+ program,


    注意事项

    本文(Chapter 2 - Control Structures.ppt)为本站会员(arrownail386)主动上传,麦多课文档分享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,请立即通知麦多课文档分享(点击联系客服),我们立即给予删除!




    关于我们 - 网站声明 - 网站地图 - 资源地图 - 友情链接 - 网站客服 - 联系我们

    copyright@ 2008-2019 麦多课文库(www.mydoc123.com)网站版权所有
    备案/许可证编号:苏ICP备17064731号-1 

    收起
    展开