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Question
| From: |
Martin Allen |
| Date: |
02-01-02 12:39 |
| Subject: |
Random number problem |
| Question: |
I am trying to make an array that contains random numbers, but each number has to be different.
I am having difficulty in writing the code. Can anyone help?
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Dim D(9) As
Integer, i As Integer, j As
Integer, k As Integer
Do Until i = 9
j = 0
Do Until j = i
k = Int((Rnd * 9 - 0 + 1) + 0)
If D(j) = k
Then
k = Int((Rnd * 9 - 0 + 1) + 0)
Else
D(i) = k
j = j + 1
End If
Loop
i = i + 1
Loop
MsgBox D(0) & D(1) & D(2) & D(3) & D(4) & D(5) & D(6) & D(7) & D(8) & D(9)
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Reply by colin harman:
is it all u want, random numbers? cos just put original x * , if u dont get wat i mean email me
Reply by Vaios:
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Dim oColRandom As New Collection
Dim D(9) As Integer
Dim k As Integer
Do Until oColRandom.Count = 9
'zero based
k = Int((Rnd * 9 - 0 + 1) + 0)
lCount = lCount + 1
On Error Resume Next
oColRandom.Add k, Chr(k)
Loop
For k = 0 To oColRandom.Count
D(k) = oColRandom.Item(k)
Next k
MsgBox D(0) & D(1) & D(2) & D(3) & D(4) & D(5) & D(6) & D(7) & D(8) & D(9) |
Answer by Martin Allen:
Thanks for that code. But I already found the answer by downloading this VB project -
http://www.vb-helper.com/HowTo/rndlist.zip.
When your code is run, the same sequence of numbers is returned - 0756318492.
Although all numbers are different, the numbers are always in the same order (unless
that's a coincidence, which I doubt). The above project sorts these numbers into a random order.
Comments
| From: |
student |
| Date: |
Sunday, December 11, 2005 at 23:54:10 |
| Comments: |
The numbers are always in the same order is because you need to declare :
RANDOMIZER TIMER
in the beginning, or else it will always give you the same set of numbers!
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| Reply: |
The statement you need is: Randomize. |
Martin Allen 1999 - 2005. Last updated Tuesday 27 December 2005 03:52:20 PM -0000.
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