Is it possible to define a bit array of custom size

Question

Is it possible to define a bit array of for example 60 bits (It's not divisible by 8)?

 bit_array = malloc(/*What should be here?*/)

All I have found defines bit arrays like

 bit_array = malloc(sizeof(long))

But this only gives 32bits (depending on architecture)

Thanks

I don't think so. malloc only has a concept of bytes. What happens at the bit level is implementation defined. — PC Luddite
– PC Luddite, Commented Oct 7, 2015 at 1:35
C has no such concept of a "bit array". You can allocate bytes. If you want 7 bits, then you'll have to allocate a whole byte and only use part of it. — Stack Exchange Broke The Law
– Stack Exchange Broke The Law, Commented Oct 7, 2015 at 1:37
If you want to use a bit field, it has to be done as part of a structure. — Andrew Henle
– Andrew Henle, Commented Oct 7, 2015 at 1:42
And to use only part of it I should use somekind of bitmask? For example in my 60bits array I should malloc 64 bits (long) and do something like myLong &= 0xFFFFFFFFFFFFFFF // It's 60 bits turned on and the other 4 off — Mr. E
– Mr. E, Commented Oct 7, 2015 at 1:46

Community · Accepted Answer · 2017-05-23 11:52:13Z

2

Here's code I wrote to manipulate bits from within an array. In my code, I allocated 60 bytes of memory from the stack which gives 480 bits for you to play with. Then you can use the setbit function to set any bit from within the 60 bytes to either a zero or one, and use getbit to find a value of a bit.

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


int getbit(unsigned char *bytes,int bit){
    return ((bytes[(bit/8)] >> (bit % 8)) & 1);
}

void setbit(unsigned char *bytes,int bit,int val){
    if (val==1){
        bytes[(bit/8)] |= (1 << (bit % 8));
    }else{
        bytes[(bit/8)] &= ~(1 << (bit % 8));
    }
}

int main(int argc, char **argv) {
    unsigned char ab[60]; // value must be the ceiling of num of bits/8
    memset(ab,0,60); // clear the whole array before use.

    //A
    setbit(ab,6,1); 
    setbit(ab,0,1); 

    //B
    setbit(ab,14,1);
    setbit(ab,9,1); 

    //C
    setbit(ab,22,1);
    setbit(ab,17,1);
    setbit(ab,16,1);

    //Change to B
    setbit(ab,16,0);

    printf("ab = %s\n",ab);
    printf("bit 17 = %d\n",getbit(ab,17));

    return 0;
}

This URL has more fragments of code for bit operations:

How do you set, clear, and toggle a single bit?

edited May 23, 2017 at 11:52

CommunityBot

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answered Oct 7, 2015 at 2:16

Mike -- No longer here

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1 Comment

Mr. E Over a year ago

Thanks a lot, this was really helpful

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