FIT100 University of Washington
Midterm #2
Friday, March 1, 2002

Sample Solution (3/8/2002)

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  Closed book, closed notes, closed neighbor, no calculators
2 points per part except as noted

Dim x As Integer

Private Sub squid()
	x=x+1+2
End Sub

Private Sub Form_Load
	x=0
	Call squid
End Sub
 

Private Sub cmdYellow_Click()

   frmFirst.BackColor = vbYellow

End Sub 

    Dim S1 as String
    Dim S2 as String    
    Dim count and Integer    
    S2 = "US"
    S1 = "Team:   "    
    count = 4    
    S2 = S1 & count    
    Msgbox(S2)

    If code1=code2 Then
         newletter = "Z"
    End If
    

    Dim dude as Integer
    dude = 0
    dude = dude + 1
    dude = 3
    dude = dude + dude
    

lblNow.Caption = TIME    

  

19.
The following two statements are legal in VB: Dim A as Integer A = 1 In a math textbook, you might also see an equation like A = 1 Is there any essential difference?  Explain briefly.
{suggested by student RF -- Thanks!  This was also a mini-quiz or CAT question, I think.} Yes, there's a difference. In math, A = 1 states a FACT -- that A and 1 are the same.  A = 1 causes nothing to change. In programming, A = 1 is an ACTION.  It causes A to take the value 1, regardless of what value it might have had before.  This is why programmers often read it aloud as "A gets 1" or "A becomes 1".

   

20.
The VB Debugger is a tool which... (mark all that apply)	___ locates program errors and informs the programmer about them. ___ automatically removes certain common errors from a program ___ lets the programmer stop execution at some desired point in the program ___ lets the programmer inspect values of variables at some desired point in the program
Last two items only: lets the programmers stop execution at a breakpoint, and inspect values of variables at that point.

  

21.
How is a "relational database" different from other things that are informally called "databases"?
See bboard post from Grace Whiteacre, 3/5/2002

 

  

22.
Which of the following is/are not a fundamental relational database operation?	___ Table ___ Union ___ Selection ___ Attribute
Table and Attribute are not operations.

 

   

23. 3 + 6 points
On a separate sheet, you have a copy of a sample solution to the basic requirements of Project 1, Part 1.	a. Suppose the program is running.  The user presses the key "3".  Which instructions (if any) of the program execute as a result?  Put a small checkmark to the left of each such instruction (leave other instructions blank).  Write directly on the handout.  b. In the ShowCodeBook procedure, there is a blank space where some lines have been omitted.  Fill in that space with code that displays the whole code book.  Note: A correct solution which uses iteration will get more credit than a correct one which doesn't.
a. Most people realized that execution flowed to the txtClearText_KeyPress event handler. Most common mistake: thinking that "3" is an error.  It's not an error from the VB point of view.  The keypresserror handler is not invoked.  Instead, the program uses an If statement to determine what to do.  It will execute ONLY one branch of the If. Another common mistake: the instructions said to check EACH instruction executed -- some people checked only some.  You still got some credit if if the intention was clear; full credit in a few cases if the omissions were inconsequential (didn't obscure the basic flow).  lines with turquoise should have been checked -- red ones should not have been. Private Sub txtClearText_KeyPress(keyascii As Integer) 'a key has been pressed in the cleartext window. 'Convert it, add it to the output On Error GoTo keypresserror Dim sletter As String 'you can do without these local variables, Dim sCodedLetter As String '... but they make the program clearer sletter = UCase(Chr(keyascii)) If sletter >= "A" And sletter <= "Z" Then sCodedLetter = colCodeBook.Item(sletter) 'look up the sletter Else sCodedLetter = sletter 'This takes care of all other visible characters 'But there is a bug here -- ' non-displayable "control characters" aren't properly handled. End If txtEncrypted.Text = txtEncrypted.Text & sCodedLetter Exit Sub keypresserror: 'Should not be needed, but is a good idea MsgBox ("lookup error on " & Chr(34) & sletter & Chr(34) _ amp; ":" & Chr(34) & keyascii & Chr(34)) sCodedLetter = sletter Resume Next Exit Sub End Sub b. See solution posted from the Project 2 web page (also was a handout in class). Tons of people were on the right track, although few got it perfectly right.  We got this interesting comment: "Do you really expect me to memorize your code?"  I answered it seriously at the time, but in retrospect perhaps the person was just joking.