创建随机颜色（System.Drawing.Color）

Question

I am tring to create random drawing colors. There is an error. Could you help me about this code.

        private Random random;

        private void MainForm_Load(object sender, EventArgs e)
        {
            random = new Random();
        }

        private Color GetRandomColor()
        {
            return Color.FromArgb(random.Next(0, 255), random.Next(0,255),random.Next(0,255));
        // The error is here
        }  

        public SolidBrush brushGet()
        {
            SolidBrush oBrush = new SolidBrush(GetRandomColor());
            return oBrush;
        }

Answer 1

This Worked for me (at the GetRandomColor):

Random random = new Random();
return Color.FromArgb((byte)random.Next(0, 255), (byte)random.Next(0, 255), (byte)random.Next(0, 255));

Answer 2

I don't see any problems with the above code, other than the Random object not being initialized before it is called to. There is also absolutely no need to initialize it in the Load event of the form; it can be done right when it's declared:

private static readonly Random Random = new Random();

Personally I'd not declare it on local scope, as far as I know you end up with the same value every time if you go about it that way. I also personally don't see the need of overcomplicating things; generating random numbers everytime and using the Color.FromAgb() method you should be fine.

Answer 3

    private Color color;
    private int b;
    public Color Random()
    {
        Random r = new Random();
        b = r.Next(1, 5);
        switch (b)
        {
            case 1:
                {
                    color = Color.Red;
                }
                break;
            case 2:
                {
                    color = Color.Blue;
                }
                break;
            case 3:
                {
                    color = Color.Green;
                }
                break;
            case 4:
                {
                    color = Color.Yellow;
                }
                break;
        }

        return color;
    }

Answer 4

The error is here

return Color.FromArgb(random.Next(0,255), random.Next(0,255), random.Next(0,255));

You need to cast random.Next(0, 255) doing this (byte)random.Next(0, 255) , and FromArgb needs 4 parameters.

Answer 5

Generating 3 random numbers and calculating the color integer can be avoided:

static Random random = new Random();

Color GetRandomColor() { return Color.FromArgb(unchecked(random.Next() | 255 << 24)); }

Color GetRandomKnownColor() {
    return Color.FromKnownColor((KnownColor)random.Next((int)KnownColor.YellowGreen) + 1); }

The unchecked | 255 << 24 unchecked | 255 << 24 part is used to avoid transparent colors.

Answer 6

我想在MainForm_Load可以创建random之前调用brushGet 。

Answer 7

using Microsoft.VisualBasic.PowerPacks;

namespace WindowsFormsApplication1
{
    public partial class Form1 : Form
    {
        Random random = new Random();

        public Form1()
        {    
            InitializeComponent();
        }

        private void ovalShape1_Click(object sender, EventArgs e)
        {        
            ovalShape1.BackStyle = BackStyle.Opaque;

            random = new Random();

            ovalShape1.BackColor = Color.FromArgb(random.Next(0, 255), random.Next(0, 255), random.Next(0, 255));

         }
     }
 }