Why is Color.IsNamedColor not working when I create a color using Color.FromArgb ()?
In my application, I allow the user to create a color and then show him the name or value of the color later. If the user selects red (full red, not red-ish), I want to show him "red". If he picks some weird color then the hex value will be fine. Here's some sample code that demonstrates the problem:
static string GetName(int r, int g, int b)
{
Color c = Color.FromArgb(r, g, b); // Note that specifying a = 255 doesn't make a difference
if (c.IsNamedColor)
{
return c.Name;
}
else
{
// return hex value
}
}
Even with very obvious colors like red IsNamedColor
, it never returns true. Looking at the ARGB values for my color and Color.Red
, I don't see any difference. However, the call Color.Red.GetHashCode()
returns a different hash code than Color.FromArgb(255, 0, 0).GetHashCode()
.
How do I create a color using user-specified RGB values and get the property Name
?
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This is probably not the fastest method, but it works. The colors don't have to exactly match the name you want to choose, for example. GetColorName(Color.FromArgb(254, 254, 0));
will still return yellow.
I deliberately ditched access modifiers
Color[] KnownColors;
void Init (){
KnownColors = (from colorInfo in typeof(Color).GetProperties(BindingFlags.Static | BindingFlags.CreateInstance |BindingFlags.Public)
where colorInfo.PropertyType == typeof (Color)
select (Color)colorInfo.GetValue(null, null)).Where (c => c.A != 0).ToArray();
}
string GetColorName(Color inColour)
{
// I'm just getting started on LINQ so im not
// sure how to do this with it (Maybe some one can help out with that)
int MinDistance = int.MaxValue;
Color closestKnown = Color.Black;
foreach (var c in KnownColors)
{
var d = ColorDistance(c, inColour);
if (d < MinDistance){
closestKnown = c;
MinDistance = d;
}
}
return closestKnown.Name;
}
int ColorDistance(Color c1, Color c2)
{
var ad = (c1.A - c2.A);
var rd = (c1.R - c2.R);
var gd = (c1.G - c2.G);
var bd = (c1.B - c2.B);
return (ad * ad) + (rd * rd) + (gd * gd) + (bd * bd);
}
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