# Convert code from unstructured to structured with embedded assignments.

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
``````
LVL 43
###### Who is Participating?

Commented:
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
``````
Matt
0

IT Services ConsultantCommented:
Thanks for looking at this Matt :)
0

Commented:
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
``````
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.
0

IT Services ConsultantCommented:
^ 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.
0

Commented:
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.
0

IT Services ConsultantCommented:
OK.  It was the wording of "immediately" that was confusing.  Thanks for clarifying what you meant.
0

EngineerAuthor Commented:
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.
0
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