Question

我正在尝试混合两种编码为整数的颜色。这是我的小功能：

int blend (int a, int b, float ratio) {
    if (ratio > 1f) {
        ratio = 1f;
    } else if (ratio < 0f) {
        ratio = 0f;
    }
    float iRatio = 1.0f - ratio;

    int aA = (a >> 24 & 0xff);
    int aR = ((a & 0xff0000) >> 16);
    int aG = ((a & 0xff00) >> 8);
    int aB = (a & 0xff);

    int bA = (b >> 24 & 0xff);
    int bR = ((b & 0xff0000) >> 16);
    int bG = ((b & 0xff00) >> 8);
    int bB = (b & 0xff);

    int A = ((int)(aA * iRatio) + (int)(bA * ratio));
    int R = ((int)(aR * iRatio) + (int)(bR * ratio));
    int G = ((int)(aG * iRatio) + (int)(bG * ratio));
    int B = ((int)(aB * iRatio) + (int)(bB * ratio));

    return A << 24 | R << 16 | G << 8 | B;
}

一切似乎都很好，但某些论据会产生错误的颜色。例如：

    int a = 0xbbccdd;
    int b = 0xbbccdd;
    int c = blend(a, b, 0.5f); // gives 0xbaccdc, although it should be 0xbbccdd

我的猜测是浮动比率或者施法都要归咎于此，但我无法弄清楚它们有什么问题......

那么在java中混合两种颜色的正确方法是什么？

Answer 1

我的猜测是，添加之后应该将int转换为int。喜欢这个

int a = (int)((aA * iRatio) + (bA * ratio));

我还建议在使用变量时使用Java命名约定。只有常数应该是上限。

Answer 2

感谢JuliusB和dARKpRINCE。我已经将它改编为接受java.awt.Color，修复了强制转换并将变量重命名为更像Java标准。它运作良好。再次感谢！

Color blend( Color c1, Color c2, float ratio ) {
    if ( ratio > 1f ) ratio = 1f;
    else if ( ratio < 0f ) ratio = 0f;
    float iRatio = 1.0f - ratio;

    int i1 = c1.getRGB();
    int i2 = c2.getRGB();

    int a1 = (i1 >> 24 & 0xff);
    int r1 = ((i1 & 0xff0000) >> 16);
    int g1 = ((i1 & 0xff00) >> 8);
    int b1 = (i1 & 0xff);

    int a2 = (i2 >> 24 & 0xff);
    int r2 = ((i2 & 0xff0000) >> 16);
    int g2 = ((i2 & 0xff00) >> 8);
    int b2 = (i2 & 0xff);

    int a = (int)((a1 * iRatio) + (a2 * ratio));
    int r = (int)((r1 * iRatio) + (r2 * ratio));
    int g = (int)((g1 * iRatio) + (g2 * ratio));
    int b = (int)((b1 * iRatio) + (b2 * ratio));

    return new Color( a << 24 | r << 16 | g << 8 | b );
}

Answer 3

感谢JuliusB，dARKpRINCE和bmauter 根据您的输入，我创建了以下功能，它将n种颜色以相等的比例混合：

public static Color blend(Color... c) {
    if (c == null || c.length <= 0) {
        return null;
    }
    float ratio = 1f / ((float) c.length);

    int a = 0;
    int r = 0;
    int g = 0;
    int b = 0;

    for (int i = 0; i < c.length; i++) {
        int rgb = c[i].getRGB();
        int a1 = (rgb >> 24 & 0xff);
        int r1 = ((rgb & 0xff0000) >> 16);
        int g1 = ((rgb & 0xff00) >> 8);
        int b1 = (rgb & 0xff);
        a += ((int) a1 * ratio);
        r += ((int) r1 * ratio);
        g += ((int) g1 * ratio);
        b += ((int) b1 * ratio);
    }

    return new Color(a << 24 | r << 16 | g << 8 | b);
}

Answer 4

@ dARKpRINCE的回答是正确的，但我有一些小技巧：

您的函数应该是静态的，因为它不依赖于任何对象字段。
通过x >>> 24代替(x >> 24) & 0xFF提取颜色的alpha分量，可以保存一个操作。
任何形式：
```
(a * (1 - ratio)) + (b * ratio)
```
可以写成：
```
a + (b - a) * ratio
```
将你需要的乘法数减半。

Answer 5

如果有人对在LibGDX中混合颜色感兴趣（基于上述解决方案，但针对LibGDX API量身定制）：

static Color blend( Color c1, Color c2, float ratio ) {
    if ( ratio > 1f ) ratio = 1f;
    else if ( ratio < 0f ) ratio = 0f;
    float iRatio = 1.0f - ratio;

    int i1 = Color.argb8888(c1);
    int i2 = Color.argb8888(c2);

    int a1 = (i1 >> 24 & 0xff);
    int r1 = ((i1 & 0xff0000) >> 16);
    int g1 = ((i1 & 0xff00) >> 8);
    int b1 = (i1 & 0xff);

    int a2 = (i2 >> 24 & 0xff);
    int r2 = ((i2 & 0xff0000) >> 16);
    int g2 = ((i2 & 0xff00) >> 8);
    int b2 = (i2 & 0xff);

    int a = (int)((a1 * iRatio) + (a2 * ratio));
    int r = (int)((r1 * iRatio) + (r2 * ratio));
    int g = (int)((g1 * iRatio) + (g2 * ratio));
    int b = (int)((b1 * iRatio) + (b2 * ratio));

    return new Color(r << 24 | g << 16 | b << 8 | a);
}

Answer 6

最简单的答案可能是：

public static Color mixColors(Color... colors) {
    float ratio = 1f / ((float) colors.length);
    int r = 0, g = 0, b = 0, a = 0;
    for (Color color : colors) {
        r += color.getRed() * ratio;
        g += color.getGreen() * ratio;
        b += color.getBlue() * ratio;
        a += color.getAlpha() * ratio;
    }
    return new Color(r, g, b, a);
}

如何正确混合两种int颜色

6 个答案: