本文介绍了生成颜色渐变在C#的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我的问题类似于,但我使用的是C#。 / p>
我有两种颜色,我有一个预定义步骤。如何检索 Color
的列表,它们是两者之间的渐变?
尝试,这不工作:
int argbMax = Color.Chocolate.ToArgb
int argbMin = Color.Blue.ToArgb();
var colorList = new List< Color>();
for(int i = 0; i {
var colorAverage = argbMin +(int)((argbMax- argbMin)* i / size);
colorList.Add(Color.FromArgb(colorAverage));
}
如果你尝试上面的代码,你会发现, code> argb 不对应于颜色的视觉渐进增加。
有任何想法吗?
解决方案
int rMax = Color(R,G,B组件)对每个单独执行相同的线性插值,然后重新组合。 。
int rMin = Color.Blue.R;
// ...并且对于B,G
var colorList = new List< Color>();
for(int i = 0; i {
var rAverage = rMin +(int)((rMax-rMin)* i / size);
var gAverage = gMin +(int)((gMax - gMin)* i / size);
var bAverage = bMin +(int)((bMax - bMin)* i / size);
colorList.Add(Color.FromArgb(rAverage,gAverage,bAverage));
}
My question here is similar to the question here, except that I am working with C#.
I have two colors, and I have a predefine steps. How to retrieve a list of Color
s that are the gradients between the two?
This is an approach that I tried, which didn't work:
int argbMax = Color.Chocolate.ToArgb();
int argbMin = Color.Blue.ToArgb();
var colorList = new List<Color>();
for(int i=0; i<size; i++)
{
var colorAverage= argbMin + (int)((argbMax - argbMin) *i/size);
colorList.Add(Color.FromArgb(colorAverage));
}
If you try the above code, you will find that a gradual increase in argb
doesn't correspond to a visual gradual increase in the color.
Any idea on this?
解决方案
You will have to extract the R, G, B components and perform the same linear interpolation on each of them individually, then recombine.
int rMax = Color.Chocolate.R;
int rMin = Color.Blue.R;
// ... and for B, G
var colorList = new List<Color>();
for(int i=0; i<size; i++)
{
var rAverage = rMin + (int)((rMax - rMin) * i / size);
var gAverage = gMin + (int)((gMax - gMin) * i / size);
var bAverage = bMin + (int)((bMax - bMin) * i / size);
colorList.Add(Color.FromArgb(rAverage, gAverage, bAverage));
}
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