strange overflow problem!

puzzling thigs in VB I am trying to figure out:

      18 * 60000 gives me back  

while   40 *1000 overflows!

Who is Participating?

Improve company productivity with a Business Account.Sign Up

VbmasterConnect With a Mentor Commented:
VB interprets the two variables as integers, and two integers gives another integer. The first expression has one variable (60000) that is a Long variable and one integer * one long => one long.

You can do something like 40 * CLng(1000) to make sure the 1000 is interpreted as a Long variable and therefore the whole expresion will be of Long type.
what data type have you declared to hold these values???
Another note.. A integer can only hold values between -32768 and +32767 (40000 is therefore too big).
Get expert help—faster!

Need expert help—fast? Use the Help Bell for personalized assistance getting answers to your important questions.

if you are using variants or you have not defined the types of variables you are using, the compiler will convert your variable to the lowest datatype that can satisfy to represent the value.

in your example:
  40 * 1000 = overflow

  40 will be considered an integer
1000 will also be considered an integer because integers range from -32,768 to 32,767. So in effect you are multiplying two integers and the result should also be an integer. Since the result is 40,000 which is bigger than the max integer value, we have an overflow.

18 * 60000
the compiler will treat the 60000 as a Long since it can represent the value.
In effect, we are multipliying longs which will also give us a Long.
the range of longs is -2,147,483,648 to 2,147,483,647. So the result is within range.

To avoid this kind of problem, make sure that you define your variables correctly.

yang, you should retract your answer, because vbmaster posted basically the same thing before you did.
Actually the easiest way to get this to work is put a # on the end of the numbers so that vb knows you're working with doubles instead of integers.

40# * 1000#

will NOT overflow. Sometimes dynamic typing helps, sometimes dynamic typing hurts.

You'ld think that VB would be smart enough to look at the *entire* equation and find the largest type and work the problem that way instead of hunting for the smallest type and generation stupid overflow errors.

If you code:

DIM Result as Double
Result = 40 * 1000

It'll still overflow even tho the receiving var is explicitly typed as a double. The right side is all integers and therefore is performed in integer math.

One really screwy thing that shows the effects of dynamic typing is:

DIM A as Double
DIM B as Double
DIM C as Double

A = 40
B = 1000
C = A * B

This will *still* overflow even though all of the vars are explicitly typed as Doubles. The assignments 'A=40' and 'B=1000' are both assigning integer values to Double Vars - and VB dutifully notices that only integers are needed and so A and B are dynamically retyped as Integers and the 'C = A * B' statement chokes.

So, the lesson is that you should make sure your numerical constants are typed the same as your vars. Using:

A = 40#
B = 1000#

will cause the compiler to treat the numerics as doubles (you can also enter 40.0 and 1000.0) and will prevent dynamic typing from biting you.

KDivad : Sorry about that. That's the price of not refreshing my browser.
yang, no problem. I've done it myself, so I always wait for the page to reload afte posting to see if it happened to me *again*. :-)
senardjiAuthor Commented:
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.