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Random Number Length

Posted on 2012-03-21
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Last Modified: 2012-08-14
When I run this code:
MsgBox (Int((rs!startnumber - rs!endnumber + 1) * Rnd + rs!endnumber) )

it returns the value of 12

When I run this code:
MsgBox (Len(Int((rs!startnumber - rs!endnumber + 1) * Rnd + rs!endnumber)))

it returns the value of 1

The len function is not returning the correct number of characters.  Why? Thanks!
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Question by:Sasha42
3 Comments
 
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by:Gustav Brock
ID: 37748565
Well, the random number isn't fixed!

Try this:

intN = Int((rs!startnumber - rs!endnumber + 1) * Rnd + rs!endnumber)
MsgBox (Len(intN))

/gustav
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Accepted Solution

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Cluskitt earned 500 total points
ID: 37748574
If the numbers are random, it could happen that it returned a number of 0-9, depending on your values.
The best way to test it is:
MsgBox (Int((rs!startnumber - rs!endnumber + 1) * Rnd + rs!endnumber)) & " (Len: " & (Len(Int((rs!startnumber - rs!endnumber + 1) * Rnd + rs!endnumber))) & ")"
which will return both the number and the length.
Also, you might want to convert the number into a string for Len. The len function is a bit quirky with integer/double:
MsgBox (Cstr(Int((rs!startnumber - rs!endnumber + 1) * Rnd + rs!endnumber))) & " (Len: " & (Len(Cstr(Int((rs!startnumber - rs!endnumber + 1) * Rnd + rs!endnumber)))) & ")"
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LVL 74

Expert Comment

by:Jeffrey Coachman
ID: 37748678
To be clear:

MsgBox (Int((rs!startnumber - rs!endnumber + 1) * Rnd + rs!endnumber) )
...returns the actual *Number*

MsgBox (Len(Int((rs!startnumber - rs!endnumber + 1) * Rnd + rs!endnumber)))
...returns the *number of characters* in the number.

So they may rarely be the same...
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