Stuck on a no brainer

ok im trying to write programs to help friends of mine with a game (they are not hacks for they game):confused:

VB Code:

Private Sub Command1_Click()
Dim a As Integer
Dim b As Double
Dim c As Double
 
a = Text1.Text
Label1.Caption = b
Label2.Caption = c
 
b = a * a * a * 8000
c = a * a * a * 8400
End Sub

I read what other people posted on similar topics and it couldnt work.
if there is help out there please... let me have it.

It might be me, but I have no idea what your after??? :)

What exactly do you want to achieve?
What should the value held in b and c be?

You may get an overflow error (depending on the value of a). Just cast one of your values to a different datatype.

b = a * a * a * CLng(8000)
c = a * a * a * CLng(8400)

what i am trying to do is write a multiplication program that does a number cubed, times 8000 and sends it to one label and cubes the same number and multiplies it by 8400 and sends it to the other label.

You need to change the datatype of variable 'a' to a Long (or Single, or Double). VB is wierd that way - I think it has to do with implicit conversion while it does "a * a * a * 8000", a becomes a value that is too big for an integer so it crashes. But don't quote me :)

Also, don't you want to set your captions AFTER you determine b and c?

This should work:

VB Code:

Private Sub Command1_Click()
    Dim a As Double
    Dim b As Double
    Dim c As Double
    
    a = Val(Text1.Text)
    
    b = a * a * a * 8000
    c = a * a * a * 8400
    
    Label1.Caption = b
    Label2.Caption = c
End Sub

thanks for the help guys. now i know how to do the other program that is a companion for this one. ;)

Quote:

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I read what other people posted on similar topics and it couldnt work.

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Does this mean you haven't actually tested the code.

have you tried using brackets

Code:

---

b = (((a * a) * a) * 8000)

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I would certainly be using Long instead of Double for variable declaration.

VB Code:

Private Sub Command1_Click()
    Dim a As Integer
    Dim b As Double
    Dim c As Double
    
    a = Val(Text1.Text)
    
    b = CDbl(a) * a * a * 8000
    c = CDbl(a) * a * a * 8400
    
    Label1.Caption = b
    Label2.Caption = c
End Sub

PS: I always test my code unless I say otherwise!
PPS: Speaking of testing code, this gives an overflow also:
b = (((a * a) * a) * 8000) :D

Eh... I know he has it figured out but wouldn't using a line such as "a^3" been much easier?