Question

我需要手动将rgba8转换为rgba5551。我从另一篇文章中找到了一些有用的代码，并希望将其修改为从rgba8转换为rgba5551。我真的没有经验，并且没有任何运气搞乱我自己的代码。

void* rgba8888_to_rgba4444( void* src, int src_bytes)
{
    // compute the actual number of pixel elements in the buffer.
    int num_pixels = src_bytes / 4;
    unsigned long*  psrc = (unsigned long*)src;
    unsigned short* pdst = (unsigned short*)src;
    // convert every pixel
    for(int i = 0; i < num_pixels; i++){
        // read a source pixel
        unsigned px = psrc[i];
        // unpack the source data as 8 bit values
        unsigned r = (px << 8)  & 0xf000;
        unsigned g = (px >> 4)  & 0x0f00;
        unsigned b = (px >> 16) & 0x00f0;
        unsigned a = (px >> 28) & 0x000f;
        // and store
        pdst[i] = r | g | b | a;
    }
    return pdst;
}

Answer 1

RGBA5551的值是它将颜色信息压缩为16位 - 或两个字节，alpha通道只有一位（开或关）。另一方面，RGBA8888对每个通道使用一个字节。（如果你不需要alpha通道，我听说RGB565更好 - 因为人类对绿色更敏感）。现在，有了5位，你得到数字0到31，所以r，g和b每个都需要转换为0到31之间的某个数字，因为它们最初是每个字节（0-255），我们乘以每个都是31/255。这是一个函数，它将RGBA字节作为输入，并输出RGBA5551为短：

short int RGBA8888_to_RGBA5551(unsigned char r, unsigned char g, unsigned char b, unsigned char a){
    unsigned char r5 = r*31/255; // All arithmetic is integer arithmetic, and so floating points are truncated. If you want to round to the nearest integer, adjust this code accordingly.
    unsigned char g5 = g*31/255;
    unsigned char b5 = b*31/255;
    unsigned char a1 = (a > 0) ? 1 : 0; // 1 if a is positive, 0 else. You must decide what is sensible.

    // Now that we have our 5 bit r, g, and b and our 1 bit a, we need to shift them into place before combining.

    short int rShift = (short int)r5 << 11; // (short int)r5 looks like 00000000000vwxyz - 11 zeroes. I'm not sure if you need (short int), but I've wasted time tracking down bugs where I didn't typecast properly before shifting.
    short int gShift = (short int)g5 << 6;
    short int bShift = (short int)b5 << 1;

    // Combine and return
    return rShift | gShift | bShift | a1;
}

您当然可以压缩此代码。

手动将rgba8转换为rgba5551

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