Get Random Color

Question

Do you know any method to generate a random Color (not a random Color Name!)?

I've already got one, but this one is'nt doing it correctly:

This only returns Green:

Random r = new Random();
BackColor = Color.FromArgb(r.Next(0, 256), r.Next(0, 256), 0);

This only returns Red:

Random r = new Random();
BackColor = Color.FromArgb(r.Next(0, 256), 0, 0);

This only returns Blue:

Random r = new Random();
BackColor = Color.FromArgb(0, 0, r.Next(0, 256));

I want my Code to return one, random Color, not only green/red/blue every time, as the above ones do.

How to solve this?

Any suggestion will be approved with joy!

Solution 1

Here's the answer I started posting before you deleted and then un-deleted your question:

public partial class Form1 : Form
{
    private Random rnd = new Random();

    public Form1()
    {
        InitializeComponent();
    }

    private void button1_Click(object sender, EventArgs e)
    {  
        Color randomColor = Color.FromArgb(rnd.Next(256), rnd.Next(256), rnd.Next(256));

        BackColor = randomColor;
    }
}

Solution 2

The original version of your last method (pre-Edit) will return all different sorts of colors. I would be sure to use a single Random object rather than create a new one each time:

Random r = new Random();
private void button6_Click(object sender, EventArgs e)
{
    pictureBox1.BackColor = Color.FromArgb(r.Next(0, 256), 
         r.Next(0, 256), r.Next(0, 256));

    Console.WriteLine(pictureBox1.BackColor.ToString());
}

It produces all sorts of different colors:

Color [A=255, R=241, G=10, B=200]
Color [A=255, R=41, G=125, B=132]
Color [A=255, R=221, G=169, B=109]
Color [A=255, R=228, G=152, B=197]
Color [A=255, R=50, G=153, B=103]
Color [A=255, R=92, G=236, B=162]
Color [A=255, R=52, G=103, B=204]
Color [A=255, R=197, G=126, B=133]

Comments

[Dmitry]

Why do you think the last code returns only Blue?

[fillobotto]

2) returns red because you are only setting the Red channel. 3) Returns blue because you are only setting the blue channel 1) Be careful that the two random you are picking probably equal each other since they are almost picked at the same time

[Rufus L]

The code in the duplicate question (not the answer) is what you are looking for

[Rufus L]

Exactly, the problem the OP was most likely having was creating a new random each time and then calling .Next, rather than re-using a global Random object.

[rschoenbach]

The above code did not work for me inside WPF project. Needed to cast the random values to a byte. Color.FromArgb((byte) rnd.Next(0, 256), (byte) rnd.Next(0, 256), (byte) rnd.Next(0, 256), (byte) rnd.Next(0, 256))