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CSE 113 Introduction to Computer Programming Lecture slides for Week 10 Monday, October 31 th, 2011 Instructor: Scott Settembre.

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Presentation on theme: "CSE 113 Introduction to Computer Programming Lecture slides for Week 10 Monday, October 31 th, 2011 Instructor: Scott Settembre."— Presentation transcript:

1 CSE 113 Introduction to Computer Programming Lecture slides for Week 10 Monday, October 31 th, 2011 Instructor: Scott Settembre

2 COURSE ADMINISTRATION Section 1 Monday, Oct. 31st, 2011 University at Buffalo: CSE 113 Instructor: Scott Settembre 2

3 For Project Assistance You can find the up-to-date hours and locations in the “Contact” section of UBLearns. Here are the names, emails, and office hours as of 10/31/2011: (Also come to additional labs if you need help) Monday, Oct. 31st, 2011 University at Buffalo: CSE 113 Instructor: Scott Settembre 3 NameEmail addressOffice hours Bich (pronounced “Bic”) Vubtvu@buffalo.eduM (11-11:50am) Bell 329 F (12-12:50pm) Bell 329 Troy Kosstroykoss@buffalo.eduTues (3-5pm) Bell 329 Scott Settembress424@buffalo.eduM (10-11am) Bell 232 F (10-11am) Bell 340 Also at all times by email AND video conference by appointment.

4 You are behind the class if… If you have not: – completed chapter 1-6, then you are behind the rest of the class. Please do the following: – Finish chapter 6 in lab this week. – Begin working on project 2. Monday, Oct. 31st, 2011 University at Buffalo: CSE 113 Instructor: Scott Settembre 4

5 Lecture and Lab this Week Lectures will go over the following: – Project 2 – switch statement – for loop Lab will have you do the following: – Finish chapter 6. – Start your project 1. – Lab Quiz on chapters 6. Monday, Oct. 31st, 2011 University at Buffalo: CSE 113 Instructor: Scott Settembre 5

6 MIDTERM / FINAL EXAM / PROJECT Section 2 Monday, Oct. 31st, 2011 University at Buffalo: CSE 113 Instructor: Scott Settembre 6

7 Midterm exam (15%) The midterm exam will be held in lecture on October 21 st. It will cover chapters 1-6.4, including the exercises in the chapter that you do in lab. It will be a series of true/false, multiple choice, and code examination (running the code in your head), similar to the quizzes that you will have taken. Grades have been released. Monday, Oct. 31st, 2011 University at Buffalo: CSE 113 Instructor: Scott Settembre 7

8 Final Exam (25%) The final exam will be during class during the last week of class. It will span two days, 45 minutes (or less) each day. It will consist of questions like the quizzes, as well as some code reading and understanding. More on this in November. Monday, Oct. 31st, 2011 University at Buffalo: CSE 113 Instructor: Scott Settembre 8

9 Project 1 Project 1 grades have been posted. Good job! Monday, Oct. 31st, 2011 University at Buffalo: CSE 113 Instructor: Scott Settembre 9

10 Project 2 Due date: November 22 nd, 2011 at or before 11:59:59 PM We will design the abstract class we will be using in class, together. The project description will be discussed on Monday and Wednesday. The abstract class and Greenfoot project you will use will be distributed in lab. Monday, Oct. 31st, 2011 University at Buffalo: CSE 113 Instructor: Scott Settembre 10

11 CHAPTER 6 Section 3 Monday, Oct. 31st, 2011 University at Buffalo: CSE 113 Instructor: Scott Settembre 11

12 The “break” command Using “break” will tell the computer to run outside of the current block (remember, a block is what is enclosed between “{“ and “}”). For example: int count=0; while (count < 5) { count=count + 1; if (count == 2) break; // when count==2, leave block } displayOutput("count == " + count); Here is the output: count == 2 Thursday, June 16th, 2011 University at Buffalo: EAS 230 Instructor: Scott Settembre 12

13 switch statment The switch statement really helps organize. switch (Age) { case 0: case 1:// If age == 0 or == 1, then newborn = 1 newborn = 1; can_drive = 0; break; case 2:// If age == 2, then execute next line case 3:// If age == 3, then toddler = 1 toddler = 1; can_drive = 0; break; case 16:// If age == 16, then execute next line case 17:// If age == 17, then child = 1, but can drive child = 1; can_drive = 1; break; default:// If age >= 18, then adult = 1 adult = 1; can_drive = 1; } Question: What happens if Age > 3 and Age < 16? Answer: The default section runs!!! Be careful with switch, you don’t always get what you think! Thursday, June 9th, 2011 University at Buffalo: EAS 230 Instructor: Scott Settembre 13

14 “break” is needed for switch..case Using “break” is especially important on the switch..case statement. For example, the “default” section will be run after the “< 18” section: int MyAge = 15; switch (MyAge < 18) { case true: displayOutput("You are a child."); default: displayOutput("You are an adult."); } displayOutput("Switch..case complete."); Here is the output: You are a child. You are an adult. Switch..case complete. Thursday, June 16th, 2011 University at Buffalo: EAS 230 Instructor: Scott Settembre 14 You would want to put a “break;” statement here. You are a child. Switch..case complete. To get this output

15 The “continue” command Using “continue” will tell the computer to restart at the top of the current looping block, performing the conditional check if needed. For example: int count=0; while (count < 3) { count = count + 1; if (count == 2) continue; // when count==2, restart block displayOutput("count == " + count); } displayOutput("Loop complete."); Here is the output: count == 1 count == 3 Loop complete. Thursday, June 16th, 2011 University at Buffalo: EAS 230 Instructor: Scott Settembre 15

16 This is a counter loop We mostly used the while statement as a “counter” loop to “count” while we did some instructions over and over. We could have also written it like this: int count=0; while (count++ < 3) { // The “++” replaces this line: count=count + 1; displayOutput("count == " + count); } displayOutput("Loop complete."); Here is the output: count == 1 count == 2 count == 3 Loop complete. Tuesday, June 11th, 2011 University at Buffalo: EAS 230 Instructor: Scott Settembre 16

17 The “for” loop Since counter loops are used so much, there is a quick and easy way to do them, the “for” loop. (sometimes called the “for..next” loop) Here is an example: for (int count=1; count <= 3; count++) { displayOutput("count == " + count); } displayOutput("Loop complete.“); Here is the output: count == 1 count == 2 count == 3 Loop complete. Tuesday, June 11th, 2011 University at Buffalo: EAS 230 Instructor: Scott Settembre 17

18 Components of a “for” loop You have the keyword “for” followed by an “expression” followed by a “statement” (or block). The expression contains 3 statements – First, a declaration or initialization that is run only once at the beginning of the loop. – Second, a Boolean expression or condition to evaluate to see if the loop should still run. – Lastly, an action to perform, usually an increment or iteration of some sort, performed at the end of the loop. “for” loops are quite nice to use as they cut down on the code clutter. Tuesday, June 28th, 2011 University at Buffalo: EAS 230 Instructor: Scott Settembre 18

19 Compare the “while” to “for” int count=0; while (count++ < 3) { displayOutput("count == " + count); } displayOutput("Loop complete."); for (int count=1; count <= 3; count++) { displayOutput("count == " + count ); } displayOutput("Loop complete." ); Tuesday, June 28th, 2011 University at Buffalo: EAS 230 Instructor: Scott Settembre 19

20 A “for” example This code finds the largest number (the Max) in an array: // Define the list and initialize values int MyList[5]={9, 2, 66, -5, 14}; // This will be the max number in the list int MaxNumber=0; // Create a loop to iterate through the list for (int x=0; x < 5; x++) { if (MyList[x] > MaxNumber) MaxNumber=MyList[x]; } // Output the max number displayOutput( MaxNumber ); Tuesday, June 28th, 2011 University at Buffalo: EAS 230 Instructor: Scott Settembre 20


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