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July9

Course: CSET 4250, Spring 2008
School: Toledo
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session *^**************************************************************************** New has begun in CSET42500412008301_Room1. Wednesday, July 9, 2008 5:55pm *^**************************************************************************** Brett Nelson--(bnelson3)>>hello David Mellick--(dmellic)>>hi David Mellick--(dmellic)>>was GA here yet? Ramya Alexander--(CSET42500412008301)>>Hi Everybody Ramya Alexander--(CSET42500412008301)>>Sorry I'm late Ramya Alexander--(CSET42500412008301)>>Hope you got the email I sent out about my log in problems David Mellick--(dmellic)>>i just got here myself Ramya Alexander--(CSET42500412008301)>>Ok then, let's discuss the midterm Ramya Alexander--(CSET42500412008301)>>I had already posted the grades earlier this week Ramya Alexander--(CSET42500412008301)>>Ok section 1 , q1 - its just the definition of syntax and semantics...all of you got this correct Ramya Alexander--(CSET42500412008301)>>q2 - b,c,d,a Ramya Alexander--(CSET42500412008301)>>q3 - %s Ramya Alexander--(CSET42500412008301)>>q4 - case1- "example of do-while outside the loop" and case 2 - "outside the loop" Ramya Alexander--(CSET42500412008301)>>because do while will have one iteration irrespective of the condition in the while Ramya Alexander--(CSET42500412008301)>>q5 - c Ramya Alexander--(CSET42500412008301)>>q6 - false Abdulhalim Al--Alshaikh--(aalalsh)>>hi Ramya Alexander--(CSET42500412008301)>>Hello Ramya Alexander--(CSET42500412008301)>>q7 - the name for constructor = same name as the class name, destructor = same name as class name with a prefix tilde (~) Ramya Alexander--(CSET42500412008301)>>constructors are used to create/intialize objects and destructors are used to delete the objects to allow the memory for reallocation Ramya Alexander--(CSET42500412008301)>>Did you all note the answers for section 1? Abdulhalim Al--Alshaikh--(aalalsh)>>yes David Mellick--(dmellic)>>yes Ramya Alexander--(CSET42500412008301)>>Moving onto section 2 Ramya Alexander--(CSET42500412008301)>>q1 - you will have to give the parse tree and left most derivation (lfd) for the expression Ramya Alexander--(CSET42500412008301)>>which goes like A = Ramya Alexander--(CSET42500412008301)>>A = Ramya Alexander--(CSET42500412008301)>> Ramya Alexander--(CSET42500412008301)>>\A = + Ramya Alexander--(CSET42500412008301)>>A = -A + Ramya Alexander--(CSET42500412008301)>>and basically expand the expr step by step Ramya Alexander--(CSET42500412008301)>>A = -A + () Ramya Alexander--(CSET42500412008301)>>A = -A + (+) Ramya Alexander--(CSET42500412008301)>>A = -A + (B+) Ramya Alexander--(CSET42500412008301)>>and so on Ramya Alexander--(CSET42500412008301)>>the last step should have the complete expression Ramya Alexander--(CSET42500412008301)>>Parse tree also has the expansion similar to the derivation..instead of writing it line by line..you draw a tree structure to represent the derivation Ramya Alexander--(CSET42500412008301)>>Now section2, q2 Abdulhalim Al--Alshaikh--(aalalsh)>>ok Ramya Alexander--(CSET42500412008301)>>[[[[A+B]1 - [C*D]2 ]3 + [([C+A]4) + [B/D]5)]6 ]7 & D]8 + [B NOT [(C+B)]9]10]11]12 Ramya Alexander--(CSET42500412008301)>>If you move from the left - you start with B, it has + and - on its either sides....and since they have the same priority, you choose the left one because of left-right associativity Ramya Alexander--(CSET42500412008301)>>so [A+B] becomes 1 Abdulhalim Al--Alshaikh--(aalalsh)>>i think i got that correct Ramya Alexander--(CSET42500412008301)>>Nobody got this one correct, few of you got it partially correct though Ramya Alexander--(CSET42500412008301)>>like the beginning part was correct and I think towards the end, you must have gotten confused Ramya Alexander--(CSET42500412008301)>>Next, you look at C which has - and * on its either side Ramya Alexander--(CSET42500412008301)>>Since * has higher priority, [c*d] becomes 2\ Ramya Alexander--(CSET42500412008301)>>then the operand [c*d] has - and + on either sides Ramya Alexander--(CSET42500412008301)>>so [a+b] + [c*d] becomes 3 Abdulhalim Al--Alshaikh--(aalalsh)>>ok Ramya Alexander--(CSET42500412008301)>>Let me call [ [a+b] + [c*d]] as P, just for future reference Ramya Alexander--(CSET42500412008301)>>Now, the subexpression - (c+a-b/d) is in parenthesis..and therefore will have the highest priority by default Ramya Alexander--(CSET42500412008301)>>so the expression inside the brackets are always executed first Ramya Alexander--(CSET42500412008301)>>Using the same rules as before , [c+a] becomes 4, [b/d] becomes 5, and (c+a-b/d) becomes 6 Ramya Alexander--(CSET42500412008301)>>Let call me (c+a-b/d) as Q for now Ramya Alexander--(CSET42500412008301)>>So now your expression is P + Q & D... Ramya Alexander--(CSET42500412008301)>>Q is the operand to look at and it has + and & on either sides and since both have same priority, (P+Q) becomes 7 Ramya Alexander--(CSET42500412008301)>>Now, D is the operand you look at and it has & and + on either sides, and both these operators have the same priority, so you choose the left one according to associativity Ramya Alexander--(CSET42500412008301)>>So (P+Q)&D becomes 8 Ramya Alexander--(CSET42500412008301)>>You just continue this to get the final answer that I mentioned above. Did you all get that? Abdulhalim Al--Alshaikh--(aalalsh)>>yes Ramya Alexander--(CSET42500412008301)>>Moving on to q3 Ramya Alexander--(CSET42500412008301)>>You will not get the desired output because the output will be "today is monday today is tuesday today is wednesday" Ramya Alexander--(CSET42500412008301)>>you will have to insert a break statement at the end of each case statement Ramya Alexander--(CSET42500412008301)>>section 2, q4 - Ramya Alexander--(CSET42500412008301)>>for pass by value - you will get value = 1, and a [] = {2, 4, 6, 8, 10} Ramya Alexander--(CSET42500412008301)>>because pass by value means that only values are passed into the function and not the variable itself, i.e a copy of the variable's value is sent into the function and so you cannot really change the value in the original variable Ramya Alexander--(CSET42500412008301)>>for Pass by Reference, value =6 and a[] = {4, 1, 2, 8, 10} Ramya Alexander--(CSET42500412008301)>>Here, the reference or address of the actual variable is sent to the function and therefore the original variable is modified Ramya Alexander--(CSET42500412008301)>>swap (value, a[0]) will swap 1 and 2 and therefore value becomes 2, and a[] = {1,4,6,8,10} Ramya Alexander--(CSET42500412008301)>>swap (a[0], a[1]) will swap 1 and 4, and therefore a [] = {4,1,6,8,10} Ramya Alexander--(CSET42500412008301)>>swap(value, a[value]) which is actually swap (2, a[2]) which swaps 2 and 6 and makes value = 6, and a[] ={4,1,2,8,10} Ramya Alexander--(CSET42500412008301)>>Did you all get that? Abdulhalim Al--Alshaikh--(aalalsh)>>ok David Mellick--(dmellic)>>think so Ramya Alexander--(CSET42500412008301)>>q5 - you will have to implement the readln function and use the predefined function 'sqr(a)' Ramya Alexander--(CSET42500412008301)>>q6 - the output is goodmorning goodnight Ramya Alexander--(CSET42500412008301)>>That is it for section 2 Ramya Alexander--(CSET42500412008301)>>Did you want discussion of the bonus questions? David Mellick--(dmellic)>>i wont be sad if we dont Abdulhalim Al--Alshaikh--(aalalsh)>>it is up to you,, if not now ,, next time Ramya Alexander--(CSET42500412008301)>>Well let me just give you the solution without explanation incase anybody wants it Abdulhalim Al--Alshaikh--(aalalsh)>>ok thanks Ramya Alexander--(CSET42500412008301)>>Position 1 ? w(def2), x(def1), y(def2), z(def1), p(def2), q(def2) Ramya Alexander--(CSET42500412008301)>>Position 2 ? w(def2), x(def1), y(def2), z(def1), p(def3), a(def3), q(def3), b(def3) Ramya Alexander--(CSET42500412008301)>>Position 3 ? w(def2), x(def1), y(def2), z(def1), p(def3), a(def4), q(def4), b(def4), c(def4) Ramya Alexander--(CSET42500412008301)>>Position 4 - w(def2), x(def1), y(def2), z(def1), p(def3), a(def3), q(def3), b(def3) Ramya Alexander--(CSET42500412008301)>>Position 5 - w(def2), x(def1), y(def2), z(def1), p(def2), q(def2) Ramya Alexander--(CSET42500412008301)>>Position 6 ? w(def1), x(def1), y(def1), z(def1) Ramya Alexander--(CSET42500412008301)>>The answer to the other bonus question - the variable 'c' Ramya Alexander--(CSET42500412008301)>>I'll be posting this chat session on the course website as well, so you can refer to it later if you want Abdulhalim Al--Alshaikh--(aalalsh)>>ok thanks Ramya Alexander--(CSET42500412008301)>>you're welcome Ramya Alexander--(CSET42500412008301)>>Did any of you have any other questions? Abdulhalim Al--Alshaikh--(aalalsh)>>thanks David Mellick--(dmellic)>>i am good Ramya Alexander--(CSET42500412008301)>>u Ramya Alexander--(CSET42500412008301)>>Ok then, we can wrap this up for today Ramya Alexander--(CSET42500412008301)>>See you next week Ramya Alexander--(CSET42500412008301)>>Bye Abdulhalim Al--Alshaikh--(aalalsh)>>see you Abdulhalim Al--Alshaikh--(aalalsh)>>byee ****************************************************************************** Session in CSET42500412008301_Room1 ended. (all participants have left). Wednesday, July 9, 2008 7:23pm ******************************************************************************

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