从NSColor中提取32位RGBA值

我有一个NSColor,我真的想要它代表的32位RGBA值.有没有什么简单的方法来获得这个,除了提取浮动组件,然后乘法和ORing,并通常做粗,依赖于endian的事情？

编辑:谢谢你的帮助.真的,我所希望的是一个Cocoa功能已经做到了这一点,但我自己很酷.

另一种更强力的方法是创建一个临时的CGBitmapContext并填充颜色.

NSColor *someColor = {whatever};
uint8_t data[4];

CGContextRef ctx = CGBitmapContextCreate((void*)data, 1, 1, 8, 4, colorSpace, kCGImageAlphaFirst | kCGBitmapByteOrder32Big);

CGContextSetRGBFillColor(ctx, [someColor redComponent], [someColor greenComponent], [someColor blueComponent], [someColor alphaComponent]);

CGContextFillRect(ctx, CGRectMake(0,0,1,1));

CGContextRelease(ctx);

FWIW,每个组件颜色值为8位没有端序问题.字节序只有16位或更大的整数.您可以以任何方式布置内存,但无论是大端还是小端机器,8位整数值都是相同的.(ARGB是Core Graphics和Core Image的默认8位格式,我相信).

为什么不呢？:

uint32_t r = (uint32_t)(MIN(1.0f, MAX(0.0f, [someColor redComponent])) * 255.0f);
uint32_t g = (uint32_t)(MIN(1.0f, MAX(0.0f, [someColor greenComponent])) * 255.0f);
uint32_t b = (uint32_t)(MIN(1.0f, MAX(0.0f, [someColor blueComponent])) * 255.0f);
uint32_t a = (uint32_t)(MIN(1.0f, MAX(0.0f, [someColor alphaComponent])) * 255.0f);
uint32_t value = (r << 24) | (g << 16) | (b << 8) | a;

然后你就知道它是如何在记忆中布局的.

或者,如果你对它更清楚:

uint8_t r = (uint8_t)(MIN(1.0f, MAX(0.0f, [someColor redComponent])) * 255.0f);
uint8_t g = (uint8_t)(MIN(1.0f, MAX(0.0f, [someColor greenComponent])) * 255.0f);
uint8_t b = (uint8_t)(MIN(1.0f, MAX(0.0f, [someColor blueComponent])) * 255.0f);
uint8_t a = (uint8_t)(MIN(1.0f, MAX(0.0f, [someColor alphaComponent])) * 255.0f);

uint8_t data[4];
data[0] = r;
data[1] = g;
data[2] = b;
data[3] = a;