What's an efficient way to convert “unsigned char” array to its “unsigned short” counterpart?

Question

What's an efficient way to convert "unsigned char" array to its "unsigned short" counterpart? I usually use the following code snippet to do so.

#define CH_LINE_PIXELS       2291
#define SCANLINE_SIZE        57301
#define CH1_INDEX            2297
#define CH2_INDEX            4592
#define CH3_INDEX            6887
#define CH4_INDEX            9182

unsigned char* pUChar = new unsigned char[SCANLINE_SIZE];

unsigned short *pUS1, *pUS2, *pUS3, *pUS4;
pUS1 = reinterpret_cast<unsigned short *>(&pUChar[CH1_INDEX]);
pUS2 = reinterpret_cast<unsigned short *>(&pUChar[CH2_INDEX]);
pUS3 = reinterpret_cast<unsigned short *>(&pUChar[CH3_INDEX]);
pUS4 = reinterpret_cast<unsigned short *>(&pUChar[CH4_INDEX]);

unsigned short us1, us2;

for (unsigned int i = 0; i < CH_LINE_PIXELS; i++) 
{   
    us1 = pUChar[CH1_INDEX + 2 * i];
    us2 = pUChar[CH1_INDEX + 2 * x + 1];
    pUS1[x] = us1 * 0x100 + us2;

    us1 = pUChar[CH2_INDEX + 2 * i];
    us2 = pUChar[CH2_INDEX + 2 * i + 1];
    pUS2[x] = us1 * 0x100 + us2;

    us1 = pUChar[CH3_INDEX + 2 * i];
    us2 = pUChar[CH3_INDEX + 2 * i + 1];
    pUS3[x] = us1 * 0x100 + us2;

    us1 = pUChar[CH4_INDEX + 2 * i];
    us2 = pUChar[CH4_INDEX + 2 * i + 1];
    pUS4[x] = us1 * 0x100 + us2;
}

Answer 1

Addressing short on byte boundary may (or may not) cause alignment issues, depending on platform.

Also, multiplying is very ineffective, why not use shifting instead? (some compilers may optimize x * 0x100, but if they don't - it's a huge performance hit when all you want is just x << 8...)

Also, as noted, reinterpret_cast may not work as you expect it to.

I would suggest, since you do assignments anyway, to copy values from array of char to a separate array of short. It costs some memory, but will save you oh so much trouble with unexpected crashes and what else.

Answer 2

Well first of all you should do:

us1 << 8 + us2 instead of multiplying by 0x100, because you want to move the first 8 bits in the upper positions, and shifting is faster then multiplication.

E.g. you have us1 = aaaaaaaa (8bits) and us2 = bbbbbbbb (another 8bits). Extending these chars to shorts would be simply padding them with 8 zeroes on the left.

Then the formula above would give you:

00000000aaaaaaaa
<< 8
aaaaaaaa00000000
+ 00000000bbbbbbbb
aaaaaaaabbbbbbbb

On the other hand you should allocate a new array of shorts for your results.