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Convert code from unstructured to structured with embedded assignments.

Posted on 2013-12-20
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Last Modified: 2014-01-18
Followup to question

http://www.experts-exchange.com/Programming/Languages/Visual_Basic/Q_28322873.html

Here is some spaghetti code with embedded assignments. I need to restructure the code without goto and line numbers.

If Abs(xi - x1) < Abs(0.00001 * xi) And Abs(yi - y1) < Abs(0.00001 * yi) Then F=1:GoTo 10
If Abs(xi - x2) < Abs(0.00001 * xi) And Abs(yi - y2) < Abs(0.00001 * yi) Then F=1:GoTo 10
If Abs(x1 - x2) < (0.0000001 * x1) Then If (yi - y1) * (yi - y2) < 0 Then F=2:GoTo 10 Else GoTo 20
If (xi - x1) * (xi - x2) < 0 Then F=2:GoTo 10
GoTo 20
10
If Abs(xi - x3) < Abs(0.00001 * xi) And Abs(yi - y3) < Abs(0.00001 * yi) Then F=F or 4:GoTo 30
If Abs(xi - x4) < Abs(0.00001 * xi) And Abs(yi - y4) < Abs(0.00001 * yi) Then F=F or 4:GoTo 30
If Abs(x3 - x4) < Abs(0.000001 * yi) Then If (yi - y3) * (yi - y4) < 0 Then F=F or 8:GoTo 30 Else GoTo 20
If (xi - x3) * (xi - x4) < 0 Then F=F or 8:GoTo 30
GoTo 20

30 F = 1

20 

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Question by:Saqib Husain, Syed
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7 Comments
 
LVL 35

Accepted Solution

by:
mvidas earned 500 total points
ID: 39731996
Hi Saqib,

Give this a shot. I used fanpages' testing sub from the other question and it all matches:
Function F(xi As Double, x1 As Double, x2 As Double, x3 As Double, x4 As Double, _
            yi As Double, y1 As Double, y2 As Double, y3 As Double, y4 As Double) As Long
 
 If (Abs(xi - x1) < Abs(0.00001 * xi) And Abs(yi - y1) < Abs(0.00001 * yi)) Or _
    (Abs(xi - x2) < Abs(0.00001 * xi) And Abs(yi - y2) < Abs(0.00001 * yi)) Then
  F = 1
 ElseIf (Abs(x1 - x2) < (0.0000001 * x1) And (yi - y1) * (yi - y2) < 0) Or _
        ((xi - x1) * (xi - x2) < 0) Then
  F = 2
 Else
  Exit Function
 End If
 
 If Not (Abs(xi - x3) < Abs(0.00001 * xi) And Abs(yi - y3) < Abs(0.00001 * yi)) And _
    Not (Abs(xi - x4) < Abs(0.00001 * xi) And Abs(yi - y4) < Abs(0.00001 * yi)) Then
  If Abs(x3 - x4) < Abs(0.000001 * yi) And Not ((yi - y3) * (yi - y4) < 0) Then Exit Function
  If Not (Abs(x3 - x4) < Abs(0.000001 * yi)) And Not ((xi - x3) * (xi - x4) < 0) Then Exit Function
 End If
 
 F = 1
End Function

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Matt
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LVL 35

Expert Comment

by:[ fanpages ]
ID: 39733343
Thanks for looking at this Matt :)
0
 
LVL 14

Expert Comment

by:Faustulus
ID: 39733359
This is what I came up with:-
Sub Restructured()

    Dim xi As Double, x1 As Double, x2 As Double, x3 As Double, x4 As Double
    Dim yi As Double, y1 As Double, y2 As Double, y3 As Double, y4 As Double
    Dim F As Integer
    
    F = Abs((Abs(xi - x1) < Abs(0.00001 * xi) And Abs(yi - y1) < Abs(0.00001 * yi)) Or _
            (Abs(xi - x2) < Abs(0.00001 * xi) And Abs(yi - y2) < Abs(0.00001 * yi)))
    If F = 0 Then
        If ((Abs(x1 - x2) < (0.0000001 * x1)) And _
           ((yi - y1) * (yi - y2) < 0)) Or _
           ((xi - x1) * (xi - x2) < 0) Then F = 2
    End If
    
    If F Then
        If (Abs(xi - x3) < Abs(0.00001 * xi) And Abs(yi - y3) < Abs(0.00001 * yi)) Or _
           (Abs(xi - x4) < Abs(0.00001 * xi) And Abs(yi - y4) < Abs(0.00001 * yi)) Then
            F = 1
        End If
        If F = 2 Then
            If ((Abs(x3 - x4) < Abs(0.000001 * yi)) And _
               ((yi - y3) * (yi - y4) < 0)) Or _
               ((xi - x3) * (xi - x4) < 0) Then F = 1
        End If
    End If
End Sub

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I point out that the original code goes to some length to assign various values to F in its lower portion all of which are converted to =1 at line 30. This renders the jump to 30 unnecessary because the value =1 can be assigned immediately.
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LVL 35

Expert Comment

by:[ fanpages ]
ID: 39733798
^ You are assuming that it is uninitialised when the code snippet commences.

If the jump to line 20 is valid during the successive tests, then setting F to 1 "immediately" (at the beginning of the code snippet) may be an inappropriate assumption.
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LVL 14

Expert Comment

by:Faustulus
ID: 39735282
That isn't what I am saying.
Since label 30 assigns F=1 and continues with label 20 any jump to label 30 can be replaced with the assignment F=1 followed by continuation at label 20.
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LVL 35

Expert Comment

by:[ fanpages ]
ID: 39735942
OK.  It was the wording of "immediately" that was confusing.  Thanks for clarifying what you meant.
0
 
LVL 43

Author Comment

by:Saqib Husain, Syed
ID: 39735991
Sorry for the confusion. I copied the code from the previous question and modified it. I forgot to delete the F=1 in after the label 30. I shall come back when I have tested the solutions. Apparently the first one seems good.
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