Why does overflow!?

This is bugging the hell out of me!

Does someone have any idea why this is chunk of code is overflowing on the 5th line!? If I remove the '1800' the overflow shifts to line 7.


iCellFrames = iCellFrames + iShiftFrames
iHours = Int(((iCellFrames / 60) / 60) / 30)
iCellFrames = iCellFrames - (108000 * iHours)
iMinutes = Int((iCellFrames / 60) / 30)
iCellFrames = iCellFrames - (1800 * iMinutes)
iSeconds = Int(iCellFrames / 30)
iCellFrames = iCellFrames - (30 * iSeconds)




I used several MsgBox inserts to check the values as it was stepping through and everything was right where it should be! I don't understand why I am getting overflow on such a horribly simple math problem!!

iCellFrames and iShiftFrames are 'Long'
iHours, iMinutes, iSeconds are 'Integer'

Many thanks for any help.

Hoag

What are the values of iCellFrames and IshiftFrames at entry to the routine?

I would put a

Debug.print iCellFrames - (1800 * iMinutes)



just before the failing instruction to see what is going on.
The computed value is out side the range of a Long.

John G

but that shouldn't be possible... being outside the range of a Long.

iCellFrames is a Long and is perfectly fine up until that point. I am then taking that value and *subtracting* from it (and it does remain above 0). Unless VB requires a Long to be above a certain number (which would be just plain stupid), this statement should go out of any bounds.

Here are a list of variable values as the block exicutes for a given value...

line 1 (iCellFrames): 154888
line 2 (iHours): 1
line 3 (iCellFrames): 46888
line 4 (iMinutes): 26

If you give me the starting values requested, I will run the routine on my computer and check out the results.


John G

you can set iShiftFrames to 108000 and iCellFrames to 46888 -- that will burn it. You can also set iShiftFrames to 1, and it will kill it as well.

'
(1800 * iMinutes) is producing your error.
Change iMinutes to a long and your problem will go away.
1800 * 26 = 46800 which exceeds the range of an Integer

John G

Sure enough, many thanks for the pointer!

Although I'm ont clear as to why VB is trying to put (1800 * iMinutes) into an Integer value, and not just doing the math. But it works now, and I know another quirk of VB... so thanks again! :)

Nick

Well 1800 will fit into an integer. An integer * integer results in an integer value also. This is why the overflow error. A long * integer results in a long. So in order to calculate this equation, either the 1800 must be declared as a long or iMinutes must be declared as a long. Since iMinutes has already be declared as an integer, the proper correction would be to type 1800 as a long. To do this you just add the type declaration character. & denotes a long, so you would put (1800& * iMinutes).

David Paulson