Java 如何正确混合两种int颜色
我尝试混合两种编码为整数的颜色。以下是我的小功能:Java 如何正确混合两种int颜色,java,colors,int,bit-manipulation,color-blending,Java,Colors,Int,Bit Manipulation,Color Blending,我尝试混合两种编码为整数的颜色。以下是我的小功能: 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); in
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;
}
我的猜测是,要么是浮点数倍增,要么是铸造,但我不知道它们有什么问题
那么,在java中混合两种颜色的正确方法是什么呢?我的猜测是,转换为int应该在添加之后进行。像这样
int a = (int)((aA * iRatio) + (bA * ratio));
我还建议在使用变量时使用Java命名约定。只有常量应该是大写 谢谢朱利叶斯和黑暗骑士。我已经对它进行了调整,以接受java.awt.Color,修复了强制转换,并将变量重命名为更像java标准的变量。它工作得很好。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 );
}
颜色混合(颜色c1、颜色c2、浮动比率){
如果(比率>1f)比率=1f;
如果(比率<0f)比率=0f,则为else;
浮动利率=1.0f-比率;
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和0xff);
inta2=(i2>>24&0xff);
int r2=((i2&0xff0000)>>16);
int g2=((i2&0xff00)>>8);
int b2=(i2和0xff);
inta=(int)((a1*iRatio)+(a2*ratio));
int r=(int)((r1*iRatio)+(r2*ratio));
int g=(int)((g1*iRatio)+(g2*ratio));
intb=(int)((b1*iRatio)+(b2*ratio));
返回新颜色(a@dARKpRINCE的答案是正确的,但我有几个小技巧:
您的函数应该是静态的,因为它不依赖于任何对象字段
通过执行x>>24
而不是(x>>24)&0xFF
来提取颜色的alpha分量时,可以保存一个操作
任何形式的:
(a * (1 - ratio)) + (b * ratio)
可以写为:
a + (b - a) * ratio
这是所需乘法次数的一半
谢谢JuliusB、dARKpRINCE和bmauter。
根据您的输入,我创建了以下函数,该函数以相等比例混合n种颜色:
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);
}
公共静态颜色混合(颜色…c){
如果(c==null | | c.length>24&0xff);
int r1=((rgb&0xff0000)>>16);
int g1=((rgb&0xff00)>>8);
int b1=(rgb和0xff);
a+=((int)a1*比率);
r+=((int)r1*比率);
g+=((int)g1*比值);
b+=((int)b1*比率);
}
返回新颜色(a以防有人对在LibGDX中混合颜色感兴趣(基于上述解决方案,但为LibGDX API定制):
静态颜色混合(颜色c1、颜色c2、浮动比率){
如果(比率>1f)比率=1f;
如果(比率<0f)比率=0f,则为else;
浮动利率=1.0f-比率;
int i1=颜色。argb888(c1);
int i2=颜色。argb888(c2);
int a1=(i1>>24&0xff);
int r1=((i1&0xff0000)>>16);
int g1=((i1&0xff00)>>8);
int b1=(i1和0xff);
inta2=(i2>>24&0xff);
int r2=((i2&0xff0000)>>16);
int g2=((i2&0xff00)>>8);
int b2=(i2和0xff);
inta=(int)((a1*iRatio)+(a2*ratio));
int r=(int)((r1*iRatio)+(r2*ratio));
int g=(int)((g1*iRatio)+(g2*ratio));
intb=(int)((b1*iRatio)+(b2*ratio));
返回新颜色(r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);
}
这很有帮助。愚蠢的我,为什么我不这样施放:/BTW。你会如何命名变量?我知道大写仅用于最终静态变量,这只是太容易阅读和定位…可能类似于“redHex”、“greenHex”和“blueHex”,因为它们代表各自颜色的十六进制值。:)那么为什么不使用ByteBuffer
来获取byte[]
数组,并从byte[]生成int值呢
array?提高性能。如果我错了,请纠正我的错误,但创建原语要比实例化对象快得多。此代码每秒必须运行数百次。我发现使用按位操作的最佳答案如下:
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);
}
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);
}