Tiago.SR Tiago.SR - 1 year ago 111
C Question

Get part of specific length of allocated memory space

I have some billions of bits loaded into RAM by the use of

- will call it big_set. I also have another amount of bits (will call it small_set) in RAM which are all set to 1 and I know its size (how many bits - I will call it ss_size), but can't predict it, as varies on each execution. ss_size can be sometimes as small as 100 or large as hundreds of millions.

I need to do some bitwise operations between small_set and some unpredictable parts of big_set of ss_size bits length. I can't just extend small_set with zeros on both most-significant and least-significant sides to make its size equal big_set's size, as that would be very RAM and CPU expensive (same operations will be done at same time with a lot of differently sized small_sets and also will do shift operations over small_set, expanding it would lead in much more bits to CPU work on).


(would be billions of bits in reality)

, so ss_size is 6. (may be an unpredictable number of bits).

I need to take 6 bits length parts of big_set, e.g.:
, etc. Obs.: not necessarily Nth 6 bits, it could be from 3rd to 9th bits, for instance. I don't know how can I get it.

I don't want to get a big_set copy with everything zeroed except the 6 bits I would be taking, like on
, as that would be also very RAM expensive.

The question is: how can I get N bits from anywhere inside big_set, so I can do bitwise operations between they and small_set? Being N = ss_size.

Answer Source

I'm not sure that the example given below will give an answer to your question, also I am not sure that the realized XOR will work correctly.

But I have tried to show how confusing can be the implementation of the algorithm, if the task is to save memory.

This is my example for case of 40 bit in big_set and 6 bit in small_set:

#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>

void setBitsInMemory(uint8_t * memPtr, size_t from, size_t to)
// sets bits in the memory allocated from memPtr (pointer to the first byte)
// where from and to are numbers of bits to be set
    for (size_t i = from; i <= to; i++)
        size_t block = i / 8;
        size_t offset = i % 8;
        *(memPtr + block) |= 0x1 << offset;

uint8_t * allocAndBuildSmallSet(size_t bitNum)
// Allocate memory to store bitNum bits and set them to 1
    uint8_t * ptr = NULL;
    size_t byteNum = 1 + bitNum / 8; // determine number of bytes for  
    ptr = (uint8_t*) malloc(byteNum);
    if (ptr != NULL)
        for (size_t i = 0; i < byteNum; i++) ptr[i] = 0;
        setBitsInMemory(ptr, 0, bitNum - 1);
    return ptr;

void printBits(uint8_t * memPtr, size_t from, size_t to)
    for (size_t i = from; i <= to; i++)
        size_t block = i / 8;
        size_t offset = i % 8;
        if (*(memPtr + block) & (0x1 << offset) )

void applyXOR(uint8_t * mainMem, size_t start, size_t cnt, uint8_t * pattern, size_t ptrnSize)
// Applys bitwise XOR between cnt bits of mainMem and pattern 
// starting from start bit in mainMem and 0 bit in pattern
// if pattern is smaller than cnt, it will be applyed cyclically
    size_t ptrnBlk = 0;
    size_t ptrnOff = 0;
    for (size_t i = start; i < start + cnt; i++)
        size_t block = i / 8;
        size_t offset = i % 8;
        *(mainMem + block) ^= ((*(pattern + ptrnBlk) & (0x1 << ptrnOff)) ? 1 : 0) << offset;
        if ((ptrnBlk * 8 + ptrnOff) >= ptrnSize)
            ptrnBlk = 0;
            ptrnOff = 0;
        if (ptrnOff % 8 == 0)
            ptrnOff = 0;

int main(void)
    uint8_t * big_set;
    size_t ss_size;
    uint8_t * small_set;
    big_set = (uint8_t*)malloc(5); // 5 bytes (40 bit) without initialization
    ss_size = 6;
    small_set = allocAndBuildSmallSet(ss_size);
    printf("Initial big_set:\n");
    printBits(big_set, 0, 39);
    // some operation for ss_size bits starting from 12th
    applyXOR(big_set, 12, ss_size, small_set, ss_size);
    // output for visual analysis
    printf("\nbig_set after XOR with small_set:\n");
    printBits(big_set, 0, 39);
    // free memory

At my PC I can see the following:

enter image description here

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