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Home/ Questions/Q 7555069
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T11:30:21+00:00 2026-05-30T11:30:21+00:00

I’ve read in a lot of places that a set data structure can be

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I’ve read in a lot of places that a set data structure can be implemented in C++ using a bit array, but I don’t fully understand this and haven’t been able to find a code sample. Does anyone have an example or a detailed explanation?

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  1. Editorial Team
    Editorial Team
    2026-05-30T11:30:23+00:00Added an answer on May 30, 2026 at 11:30 am

    Using a bit field to implement a set only works if there are only a few possible elements that can go in the set, because you need a bit for each of them. A bit-set of all 32-bit integers, for instance, would need 2^32 bits, or about 500 megabytes.

    The good news is, if there are few enough possible elements that space isn’t a problem, it’s really, really fast.

    What you do, essentially, is define a bit array such that each bit corresponds to one possible element. Each bit corresponding to an element that’s in the set is 1; the others are 0.

    Will post sample C code in a bit (no pun intended). I think C++ may offer direct library support for bit-sets, but unfortunately I don’t speak it.

    EDIT: The following sample code, which I just wrote, is for a bit-set which can contain the numbers 0 through 31. Allowing support for an arbitrary number of elements would be significantly more complicated, though certainly useful.

    #include <stdint.h>
    #include <stdio.h>
    
    const BS_SIZE = 32;
    typedef uint32_t bitset32;
    
    /* Takes a pointer to a bit set and an int from 0 to 31,
     * and adds the int to the bit set.
     */
    void add_element(bitset32 *bs, int elt)
    {
        *bs |= (1 << elt);
    }
    
    /* Takes a pointer to a bit set and an int from 0 to 31,
     * and removes the int from the bit set.
     */
    void remove_element(bitset32 *bs, int elt)
    {
        *bs &= ~(1 << elt);
    }
    
    /* Takes a pointer to a bit set and an int from 0 to 31,
     * and returns 1 if the int is in the bit set, 0 otherwise.
     */
    int has_element(bitset32 *bs, int elt)
    {
        return (*bs >> elt) & 1;
    }
    
    /* Takes a pointer to a bit set and prints each element in it. */
    void print_all_elements(bitset32 *bs)
    {
        bitset32 bits = *bs;
        int i;
        for (i = 0; i < BS_SIZE; i++) {
            if (bits & 1) {
                printf("%d\n", i);
            }
            bits >>= 1;
        }
    }
    
    /* Some test code. Prints:
     * 0 in test
     * 0
     * 20
     * 31
     */
    int main()
    {
        bitset32 test = 0;
        add_element(&test, 0);
        add_element(&test, 13);
        add_element(&test, 13);
        add_element(&test, 20);
        add_element(&test, 28);
        remove_element(&test, 13);
        remove_element(&test, 20);
        remove_element(&test, 28);
        if (has_element(&test, 0)) {
            printf("0 in test\n");
        }
        if (has_element(&test, 20)) {
            printf("20 in test\n");
        }
        add_element(&test, 20);
        add_element(&test, 31);
        print_all_elements(&test);
        return 0;
    }
    
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