
Down Rodeo said: Ah, um, yes? :p
I'm sure it makes sense. Your code looks extensible as well, for instance, the prime numbers could
be replaced by a more complete list. If you restructure a bit you can remove the goto.
If I restructeireor it a bit it would look like
code Public Function calcformula()
a = INa.Text
b = INb.Text
c = INc.Text
radican = (b * b)  4 * (a * c) 'lets hope you computer knows oder of operations
If radican < 0 Then 'the radican is less than 0 see summary
radican = Abs(radican) 'get the absolute value of the radican
I = True 'remember to inlcude the I in the final answer
End If
x1 = (1 * b) + Sqrt(radican) 'radican should now be positive and not result in 0
x2 = (1 * b)  Sqrt(radican) 'radican should now be positive and not result in 0
divisor_x1 = 2 * a
divisor_x2 = 2 * a
remainder = IEEERemainder(x1, divisor_x1)
If remainder = 0 Then
x1 = x1 / divisor_x1
End If
remainder = IEEERemainder(x2, divisor_x2)
If remainder = 0 Then
x2 = x2 / divisor_x2
End If
Return x1 And x2 And divisor_x1 And divisor_x2
End Function
'''<summary>
''' the radican_check is meant to abort getting the square root of the radican
''' beause the radican is less than 0 meaning √(radican) if you remember radicals
''' then you know that you cannot get the square root of a negative number
''' this is not equaviliant to √(radical) that is possible
''' so you need to drop the negative and include an i (i as imaginary number) in front
''' of the radical thus i√(radican) however Visual Basic does not account for that
''' and does not throw an error (as they do when you try to divide by 0)
''' it simply returns 0 (not null or nothing) thus you will get 0/denominator
''' where denominator is the 2*A and 0 is x1 (ie if A=1 B=24 C=144 the end result would be 12/2 because (1*b) is not inside the √())
''' it is not 24/2 because below that we divide the numerator by the denominator (24/2 = 12/2)
'''</summary>
Shitty ass hotel wireless internet!!!
It's so slow it posted 31 times!!!!



≡

2009 Dec 22 at 10:06 PST
— Ed. 2009 Dec 22 at 10:11 PST




UPDATE
code Public Function calcformula()
a = INa.Text
b = INb.Text
c = INc.Text
discriminant = (b * b)  4 * (a * c) 'lets hope you computer knows oder of operations
If discriminant < 0 Then 'the discriminant is less than 0 see summary
discriminant = Abs(discriminant) 'get the absolute value of the discriminant
I = True 'remember to inlcude the I in the final answer
End If
discriminant_remainder = Sqrt(discriminant)
If discriminant_remainder = 0 Then
x1 = (1 * b) + Sqrt(discriminant) 'discriminant should now be positive and not result in 0
x2 = (1 * b)  Sqrt(discriminant) 'discriminant should now be positive and not result in 0
divisor_x1 = 2 * a
divisor_x2 = 2 * a
remainder = IEEERemainder(x1, divisor_x1) ' have to check if the complex numbers are functioning correctly
If remainder = 0 Then ' have to fully reduce numbers
x1 = x1 / divisor_x1 ' if cannot solve for square root and NOT get decimal leave as √() with I if applicable
divisor_x1 = divisor_x1 / divisor_x1
End If
remainder = IEEERemainder(x2, divisor_x2)
If remainder = 0 Then
x2 = x2 / divisor_x2
divisor_x2 = divisor_x2 / divisor_x2
End If
End If
If discriminant_remainder <> 0 Then
b = (1 * b)
Return b And discriminant
ElseIf errorMsg = "" Then
Return x1 And x2 And divisor_x1 And divisor_x2
Else
Return errorMsg
End If
End Function
'''<summary>
''' the discriminant check (Function check()) is meant to abort getting the square root of the discriminant
''' beause the discriminant is less than 0 meaning √(discriminant) if you remember radicals
''' then you know that you cannot get the square root of a negative number
''' this is not equaviliant to √(radical) that is possible
''' so you need to drop the negative and include an i (i as imaginary number) in front
''' of the radical thus i√(discriminant) however Visual Basic does not account for that
''' and throws the error NaN (Not a Number) thus you will get 0/denominator
''' where denominator is the 2*A and 0 is x1 (ie if A=1 B=24 C=144 the end result would be 12/2 because (1*b) is not inside the √())
''' it is not 24/2 because below that we divide the numerator by the denominator (24/2 = 12/2)
'''</summary>
code Public Function check()
If Not I = True Then ' If Not I=True must be here first because if I is true than we changed the value to get the correct square root so the rest of the function would be corrupt
If discriminant < 0 Then ' If the discriminant is less than zero then there should be 1 repeated real number check if that is true
If x1 <> x2 Then errorMsg = "Something went wrong" 'if x1 does not equal x2 then something went wrong report it
End If
If discriminant > 0 Then ' If the discriminant is greater than zero then there should be 2 different real numbers check if that is true
If x1 = x2 Then errorMsg = "Something went wrong" 'if x1 does equal x2 then something went wrong report it
End If
End If
If I = True Then
End If
If errorMsg = "" Then
Return 0
Else
Return errorMsg
End If
End Function
code Private Sub Bcalculate_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Bcalculate.Click
If EquationToolStripMenuItem.Checked = True Then
calcequation()
ElseIf FormulaToolStripMenuItem.Checked = True Then
calcformula()
End If
If I = True Then
OUTx.Text = x1.ToString() + "/" + divisor_x1.ToString() + "i" + " " + x2.ToString() + "/" + divisor_x2.ToString() + "i"
ElseIf errorMsg <> "" Then
OUTx.Text = errorMsg
ElseIf discriminant_remainder <> 0 Then
OUTx.Text = b.ToString() + " + " + "√(" + discriminant.ToString() + ")"
Else
OUTx.Text = x1.ToString() + "/" + divisor_x1.ToString() + " " + x2.ToString() + "/" + divisor_x2.ToString()
End If
End Sub
All I have to do is add something to determine if it is actually
a quadratic and return a value if it is not.



≡

2009 Dec 22 at 11:35 PST
— Ed. 2009 Dec 22 at 11:44 PST



Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧

Well, I'm very impressed, your code is a whole lot cleaner and goddamn if it doesn't check itself for errors as well. On top of that it seems to work. Well done sah!



≡

2009 Dec 22 at 17:14 PST



SRAW
Rocket Man
2007 Nov 6 • 2525
601 ₧

what the hell is a radican?



≡

2009 Dec 22 at 19:50 PST



SRAW
Rocket Man
2007 Nov 6 • 2525
601 ₧

and why would you need to check if it is a quad equation if the guy just inputs a, b and c? or just check if the guy inputs a string...................................................................................................................................................................................................................................................................................................................................................................................................................?>?>?>?>?>>>?>?>?>?>??>?>?>?>??>?>?>?



≡

2009 Dec 22 at 19:52 PST



Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧

The radican is the bit being squarerooted in a square root.
SRAW said: and why would you need to check if it is a quad equation if the guy just inputs a, b and c?
Because of the midgets. No, if the user puts a = 0 then it's not a quadratic.



≡

2009 Dec 23 at 02:21 PST




I am also going to put a nice little bit in there to put into the equation to see if it equals 0.
And apparently there is an alternative quadratic equation with C on top?



≡

2009 Dec 23 at 06:34 PST
— Ed. 2009 Dec 23 at 06:35 PST



Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧




≡

2009 Dec 23 at 06:53 PST




Down Rodeo said:
So A can be zero? if you use the alt form?
Now its scientific notation time.
Oh no its polynomial time.
I jus' realized I am putting in all these auxilairy features of
the calculator and it cannot even add or subtract!
Thats right I haven't even coded it to do basic functions.
My main problem was that I had 1 inputbox [because 2 is jus' gay]
and I didn't want to have to go through the problem of separating
functions [ie 43+18 would be 4  3 + 18 5 different strings/ints]
then I realized I could jus' get the value of the inputbox
when the pressed a function.
So lets say they input 43 then they hit minus I could get the
current value of the inputbox and then add the + then get the
current input of the inputbox again compare it with the previous
one and remove any similarities [so that 43 =? 43+ would result in
43 and +] with a few tweaks and adjustments it would work even
better!



≡

2009 Dec 23 at 07:39 PST
— Ed. 2009 Dec 23 at 08:00 PST



Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧

If A is zero it's just a line.



≡

2009 Dec 23 at 08:23 PST




Oh, I also want to put in a nice little demonstration of how to
correctly use the quadratic formula. The thing is though its pretty
bland without audio, and I am not one to narrate. So DR you want to
narrate on how to do some quadratics for me?



≡

2009 Dec 23 at 12:45 PST
— Ed. 2009 Dec 23 at 12:46 PST



Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧

Not particularly, I'm very microphone shy. I've also been told I mumble.



≡

2009 Dec 23 at 13:01 PST




Then who wants to narrate?



≡

2009 Dec 23 at 14:14 PST




I got a microphone ill do it if need be.But you gatta type out what i need to say lol. If u need me too my email is triple_edgesk@yahoo.com or just tell me in general section cause i dont check here often.
Make awkward sexual advances, not war. Down Rodeo said: Dammit, this was the one place that didn't have this, but noooooo, molkman pisses all over that



≡

2009 Dec 23 at 17:50 PST
— Ed. 2009 Dec 23 at 17:54 PST



SRAW
Rocket Man
2007 Nov 6 • 2525
601 ₧

dont ask havokk! she has a girly voice, just use the microsoft speech thingy



≡

2009 Dec 23 at 22:36 PST




Down Rodeo said: Well, I'm very impressed, your code is a whole lot cleaner and goddamn if it doesn't check itself for errors as well. On top of that it seems to work. Well done sah!
I should have put in a statement or expression that checks the calculation again and fixed any errors. Also, I should have check, constantly, for overflow errors.
And, if it truly was user input error I should have had a try{} catch{} statement to handle that.
Moreover, I could have had several catches (not to be confused with the try{} catch{}) to find out where, if anything, something went wrong; and, possibly fix it, or at least, report exactly what went wrong.
Lastly, I should have included some kind of error reporting.



≡

2010 Jan 23 at 15:41 PST
— Ed. 2010 Jan 23 at 19:59 PST




So....How was the audio?
Make awkward sexual advances, not war. Down Rodeo said: Dammit, this was the one place that didn't have this, but noooooo, molkman pisses all over that



≡

2010 Jan 24 at 22:47 PST




I never finished the project. I listened to the audio, and it wasn't bad. I may have you do more audio for me in the future, if you're up to it.



≡

2010 Jan 24 at 23:50 PST
— Ed. 2010 Jan 24 at 23:50 PST




sprinkles said: I never finished the project. I listened to the audio, and it wasn't bad. I may have you do more audio for me in the future, if you're up to it.
I'm up for it. I wanna get used to talking into a mic anyways.
Make awkward sexual advances, not war. Down Rodeo said: Dammit, this was the one place that didn't have this, but noooooo, molkman pisses all over that



≡

2010 Jan 25 at 14:01 PST




Good in a few days I will have a project for you.



≡

2010 Jan 25 at 18:07 PST



