Visual Basic - Guess the colour program

Question

I am trying to make a small game in which the computer gives a random code of colours (red green yellow and blue) and then you must try and guess them... I am having trouble making the colours random though.

The colours are the backcolour of 4 buttons. The code is four colours long. The player then clicks on some buttons just below that and tries to guess the code. Each click changes the colour once. If the player guesses the correct colour which is in the correct place then the colour is revealed.

So far I have this:

(Problem is : Overload resolution failed because no Public '=' can be called with these arguments:'Public Shared Operator =(left As System.Drawing.Color, right As System.Drawing.Color) As Boolean': Argument matching parameter 'right' cannot convert from 'Integer' to 'Color'. (See below REM color 1 to REM End) Problem was generated by the computer , program written in Visual Basic , windows form application)

Dim turn = 0
Dim generator As New Random
Dim color1 = generator.Next(1, 4)
Dim color2 = generator.Next(1, 4)
Dim color3 = generator.Next(1, 4)
Dim color4 = generator.Next(1, 4)


Private Sub NewToolStripMenuItem_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles NewToolStripMenuItem.Click

    REM color 1                 
    If color1 = 1 Then              
        color1 = Color.Red
    End If
    If color1 = 2 Then
        color1 = Color.Blue
    End If
    If color1 = 3 Then
        color1 = Color.Yellow
    End If
    If color1 = 4 Then
        color1 = Color.Green
    End If
    REM color 2
    If color2 = 1 Then
        color2 = Color.Red
    End If
    If color2 = 2 Then
        color2 = Color.Blue
    End If
    If color2 = 3 Then
        color2 = Color.Yellow
    End If
    If color2 = 4 Then
        color2 = Color.Green
    End If
    REM color 3
    If color3 = 1 Then
        color3 = Color.Red
    End If
    If color3 = 2 Then
        color3 = Color.Blue
    End If
    If color3 = 3 Then
        color3 = Color.Yellow
    End If
    If color3 = 4 Then
        color3 = Color.Green
    End If
    REM color 4
    If color4 = 1 Then
        color4 = Color.Red
    End If
    If color4 = 2 Then
        color4 = Color.Blue
    End If
    If color4 = 3 Then
        color4 = Color.Yellow
    End If
    If color4 = 4 Then
        color4 = Color.Green
    End If
  REM End
    Button1.BackColor = color1
    Button2.BackColor = color2
    Button3.BackColor = color3
    Button4.BackColor = color4

Answer 1

Thanks for your help but I found some help from another guy , Clint of "Coding4fun.com". Work properly now. What I had done was as yenthefirst had said was confused the computer by calling color1 for example an integer then asking for a color. What I have done now was just as BlueMonkMN wich was to begin a case ie.

Private Function getrandomcolour() As Color
    Select Case generator.[Next](1, 6)
        Case 1
            Return Color.Red
            Exit Select
        Case 2
            Return Color.Green
            Exit Select
        Case 3
            Return Color.Yellow
            Exit Select
        Case 4
            Return Color.Blue
            Exit Select
        Case 5
            Return Color.Pink
            Exit Select
        Case 6
            Return Color.Turquoise
            Exit Select
    End Select
    If Button1.BackColor = Button2.BackColor Then
        getrandomcolour()
    End If
    If Button3.BackColor = Button4.BackColor Then
        getrandomcolour()
    End If
    If Button1.BackColor = Button3.BackColor Then
        getrandomcolour()
    End If
    If Button2.BackColor = Button3.BackColor Then
        getrandomcolour()
    End If
End Function

Thanks for your lovely answers anyway!

Answer 2

When you Dim color1 = generator.Next(1, 4), that implicitly defined color1 as an integer because that's what generator.Next returns. I suggest (for your own benefit and that of anyone reading your code) that you use "As" to explicitly define the type of your variables.

It looks like you're trying to use the same color1 variable to contain both the integer and enumerated ("Color") version of your color. That's going to get quite confusing because enumerated values can also be directly converted to and from integers if you explicitly force that to happen, and then you won't know whether a particular value represents your own color coding system or the system's. I suggest that you create a function that generates 1 of 4 random colors like this:

Function GenerateColor() As Color
    Select Case generator.Next(1,5)
       Case 1
          Return Color.Red
       Case 2
          Return Color.Blue
       Case 3
          Return Color.Yellow
       Case 4
          Return Color.Green
    End Select
End Function

And use that function to initialize your random values.

Answer 3

The problem lies in that you are setting colorN (an integer) to a Color object.

As an addendum to the excellent answers already here...

generator.Next(1,4)

will only return values between 1 and 3!

What you really want is:

generator.Next(1,5)

Answer 4

color1,color2,color3,color4 are integers.

when you write: Dim color1 = generator.Next(1, 4)

generator.Next produces an integer, so color1 is defined as an integer. When you write color1 = Color.Red

it's confused, because you're trying to set a Color to a variable which is an integer.

There's cleaner ways to write the code in general, but you could probably do something like this:

option explicit
Dim color_index as Integer
Dim color1 as System.Drawing.Color
color_index=generator.Next(1,4)
[choosing logic]
color1 = Color.red
[blah blah blah]
Button.backcolor = color1

Additional note: as the first answer says, there's really a better way to write this whole thing. The following pseudo-code is how I'd approach the problem: (sorry, I don't remember exact VB syntax off the top of my head)

Button_Array = [Button1,Button2,Button3,Button4]
Color_array= [Color.red,Color.blue,Color.green,Color.yellow]
for i in 1 to 4
   color_index=generator.Next(1,4)
   Button_Array[i].backcolor = Color_array[color_index]
endfor

That's much more readable, and more maintainable. (consider how many lines you have to add if you want to add a 5th button, or a 5th color, in your original listing, or my pseudo-code listing)

Answer 5

You should build a dictionary of colors like this:

private _Generator as Random
private _Colors as Dictionary(of Integer, Color)
private _Color1 as Color
private _Color2 as Color
private _Color3 as Color
private _Color4 as Color


Public Sub New()
    _Generator = new Random()
    _Colors.Add(1, Colors.Red)
    _Colors.Add(2, Colors.Blue)
    _Colors.Add(3, Colors.Yellow)
    _Colors.Add(4, Colors.Green)    
End Sub

Private Sub NewToolStripMenuItem_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles NewToolStripMenuItem.Click

    _Color1 = _Colors(_Generator.Next(1, 4))
    _Color2 = _Colors(_Generator.Next(1, 4))
    _Color3 = _Colors(_Generator.Next(1, 4))
    _Color4 = _Colors(_Generator.Next(1, 4))

    Button1.BackColor = _Color1
    Button2.BackColor = _Color2
    Button3.BackColor = _Color3
    Button4.BackColor = _Color

End sub

Answer 6

Arrays in Visual Basic

I've no idea what your specific problem is, but that massive switch should be just an array of colours. If you can't compare colours with = for some reason, make the code an array of integers, and work in integers, and look up the colours in an array when you need to display them.