Android的颜色变深 [英] Android Color Darker

问题描述

好了，所以我在我的应用程序中的整型变量。这是一个色彩的价值，在我的preferences被设置由一个颜色选择器。现在，我需要同时使用的颜色和任何颜色它可能是一个黑暗的版本。

Okay, so I have an integer variable in my application. It's the value of a color, being set by a color picker in my preferences. Now, I need to use both that color and a darker version of any color it might be.

现在我知道标准Java有一个Color.darker（）方法，但似乎并没有在Android的等效。有谁知道的等效或任何变通办法？谢谢你。

Now I know in standard Java there is a Color.darker() method, but there doesn't seem to be an equivalent in Android. Does anyone know of an equivalent or any workarounds? Thanks.

推荐答案

最简单的，我认为，将转换为HSV，做变黑那里，并转换回：

The easiest, I think, would be to convert to HSV, do the darkening there, and convert back:

float[] hsv = new float[3];
int color = getColor();
Color.colorToHSV(color, hsv);
hsv[2] *= 0.8f; // value component
color = Color.HSVToColor(hsv);

（这是未经测试）。为了减轻由东西> 1.0乘以值部件。你得的结果夹在区间[0.0，1.0]。你也可以尝试使用其他美白配方，因为只需乘不会淡化黑眼圈。

(This is untested.) To lighten, multiply the value component by something > 1.0. You'll have to clamp the result to the range [0.0, 1.0]. You might also experiment with other lightening formulas, since simply multiplying isn't going to lighten black.

编辑：其实，减仓，比一个更好的策略上面可能会减少1.0值部件的区别：

Actually, to lighten, a better strategy than the above might be to reduce the difference from 1.0 of the value component:

hsv[2] = 1.0f - 0.8f * (1.0f - hsv[2]);

这是完全平行为变暗的方法，只是用1作为代替0它工作以减轻任何颜色（甚至黑色），并且不需要任何夹紧原点。它可以简化为：

This is entirely parallel to the approach for darkening, just using 1 as the origin instead of 0. It works to lighten any color (even black) and doesn't need any clamping. It can be simplified to:

hsv[2] = 0.2f + 0.8f * hsv[2];

然而，由于浮点运算可能四舍五入的影响，我会稍微关注的是，结果可能超过1.0F（由也许有位）。

However, because of possible rounding effects of floating point arithmetic, I'd be slightly concerned that the result might exceed 1.0f (by perhaps one bit).

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