逐字节读取内存

Question

I've got a long[] which I'd like to read two bits at a time. My data are the binary numbers 00,01,10, and 11 concatenated end to end, stuffed in a long, and then stuffed in an array.

I'll be reading a long stretch of this data at once, possibly starting halfway through, and it seems like it would make more sense to read straight from memory, two bytes at a time, rather than iterating through the long[] and pulling two bits at a time with a mask.

I can't seem to figure out how I'd go about this, and I've never been great with directly accessing memory (since I was brought up on java).

I've tried instantiating an array

unsigned long t[5];
t[0] = 4294967295;
t[1] = 0;
t[2] = 4294967294;
t[3] = 4294967296;
t[4] = 1;

and then printing *(&t) and *(&t+1) , but the plus one of course knows that it's the size of long and goes adds the appropriate value.

Answer 1

Use a pointer to a byte-sized data type. Try this:

unsigned char* p = (char *)t;

And use p pointer.

Answer 2

Well, you could always cast a pointer to char*, to get it byte by byte, if you also need to actually read it 2-bytes at a time, you'll need to use & and a proper bitmask (ie 0x3 to get the lowermost 2 bits). If you want, you can always just shift this mask with the << or >> operators to match further up or down in your current byte.

Answer 3

Use an std::vector<bool> instead of a long[] for this. Retrieving the i 'th 2-bit block can be done using

(int(v[i*2]) << 1) | v[i*2+1]

Answer 4

Some simple arithmetic:

for (unsigned int i = 0; i != 5; ++i)
{
     for (unsigned int k = 0; k != CHAR_BIT * sizeof(unsigned long int); k += 2)
     {
         std::printf("%02X ", n % 3);  // print last two bits
         n /= 4;                       // move down by two bits
     }
}

Instead of 5 you can of course say sizeof(t)/sizeof(t[0]) , if that's an option, or you can use iterators std::begin(t) / std::end(t) .

Answer 5

Cast the array to unsigned char * and print that. The first two bytes (endianness aside) would then be:

((unsigned char *)t)[0]
((unsigned char *)t)[1]

...and so on

Edit: incidentally, t automatically decays to a pointer on its own, there's no need to do &t , you're just confusing yourself further.