CODEWITHSUNDEEP
arraysalgorithm
Adam
Adam

Reputation: 41

String to very long sequence of length less than 1 byte

I can't guess how to solve following problem. Assume I have a string or an array of integer-type variables (uchar, char, integer, whatever). Each of these data type is 1 byte long or more.

I would like to read from such array but read a pieces that are smaller than 1 byte, e.g. 3 bits (values 0-7). I tried to do a loop like

cout << ( (tab[index] >> lshift & lmask) | (tab[index+offset] >> rshift & rmask)  );

but guessing how to set these variables is out of my reach. What is the metodology to solve such problem?

Sorry if question has been ever asked, but searching gives no answer.

Upvotes: 1

Views: 116

Answers (3)

Flayneorange
Flayneorange

Reputation: 228

Here's the best method I could come up with for setting individual bits of a variable: Assume we need to set the first four bits of variable1 (a char or other byte long variable) to 1010

variable1 &= 0b00001111; //Zero the first four bytes
variable1 |= 0b10100000; //Set them to 1010, its important that any unaffected bits be zero

This could be extended to whatever bits desired by placing zeros in the first number corresponding to the bits which you wish to set (the first four in the example's case), and placing zeros in the second number corresponding to the bits which you wish to remain neutral in the second number (the last four in the example's case). The second number could also be derived by bit-shifting your desired value by the appropriate number of places (which would have been four in the example's case).

In response to your comment this can be modified as follows to accommodate for increased variability:

For this operation we will need two shifts assuming you wish to be able to modify non-starting and non-ending bits. There are two sets of bits in this case the first (from the left) set of unaffected bits and the second set. If you wish to modify four bits skipping the first bit from the left (1 these four bits 111 for a single byte), the first shift would be would be 7 and the second shift would be 5.

variable1 &= ( ( 0b11111111 << shift1 ) | 0b11111111 >> shift2 );

Next the value we wish to assign needs to be shifted and or'ed in. However, we will need a third shift to account for how many bits we want to set. This shift (we'll call it shift3) is shift1 minus the number of bits we wish to modify (as previously mentioned 4).

variable1 |= ( value << shift3 );

Upvotes: 0

angelatlarge
angelatlarge

Reputation: 4150

I am sure this is not the best solution, as there some inefficiencies in the code that could be eliminated, but I think the idea is workable. I only tested it briefly:

void bits(uint8_t * src, int arrayLength, int nBitCount) {
   int idxByte = 0;      // byte index
   int idxBitsShift = 7; // bit index: start at the high bit
   // walk through the array, computing bit sets
   while (idxByte < arrayLength) {
       // compute a single bit set
       int nValue = 0;
       for (int i=2; i>=0; i--) {
        nValue += (src[idxByte] & (1<<idxBitsShift)) >> (idxBitsShift-i);
        if ((--idxBitsShift) < 0) {
            idxBitsShift=8;
            if (++idxByte >= arrayLength) 
                break;
        }
       }
       // print it
       printf("%d ", nValue);
   }

}

int main() {
    uint8_t a[] = {0xFF, 0x80, 0x04};
    bits(a, 3, 3);
}

The thing with collecting bits across byte boundaries is a bit of a PITA, so I avoided all that by doing this a bit at a time, and then collecting the bits together in the nValue. You could have smarter code that does this three (or however many) bits at a time, but as far as I am concerned, with problems like this it is usually best to start with a simple solution (unless you already know how to do a better one) and then do something more complicated.

Upvotes: 1

pgf
pgf

Reputation: 65

In short, the way the data is arranged in memory strictly depends on :

  • the Endianess
  • the standard used for computation/representation ( usually it's the IEEE 754 )
  • the type of the given variable

Now, you can't "disassemble" a data structure with this rationale without destroing its own meaning, simply put, if you are going to subdivide your variable in "bitfields" you are just picturing an undefined value.

In computer science there are data structure or informations structured in blocks, like many hashing algorithms/hash results, but a numerical value it's not stored like that and you are supposed to know what you are doing to prevent any data loss.

Another thing to note is that your definition of "pieces that are smaller than 1 byte" doesn't make much sense, it's also highly intrusive, you are losing abstraction here and you can also do something bad.

Upvotes: 0

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