Re[3]: [Dev-C++] using char* as bitfield

First:
sizeof(mybits) is just size of pointer - not size of the array in bytes or
bits.

mybits[bit/8] |=3D 1 << (bit&3);

isn't so complicated. Each character contains 8 bits. So by dividing bit
with 8, I get which character the bit is stored in.

bit&3 will use the 3 least significant bits for a numeric range of 0-7.

1 << (bit&3) will then access one of the 8 bits in the character.

char test_bit(const char* mybits,unsigned bit) {
    return (mybits[bit/8] | (1 << (bit&3))) !=3D 0;
}

char clear_bit(const char *mybits,unsigned bit) {
    mybits[bit/8] &=3D ~(1 << (bit&3));
}

/Per W

On Tue, 30 Sep 2003, Christian Sch=E4fer wrote:

> hi Per,
>
> I'll show you my code now. somehow it does not work as planned..
>
> char *alloc_bits(unsigned int);
> void set_bit(char*, unsigned int);
>
> int main(int argc, char *argv[])
> {
>     char *mybits;
>
>     mybits =3D alloc_bits(60);
>
>     if (mybits) printf("bits[%d]: %X\n",sizeof(mybits), *mybits);
>
>     if (mybits) set_bit(mybits,2);
>
>     if (mybits) printf("bits[%d]: %X\n",sizeof(mybits), *mybits);
>
>     system("PAUSE");
>     return 0;
> }
>
> char *alloc_bits(unsigned int bits)
> {
>     return calloc(((bits+7)/8), sizeof(char));
> }
>
> void set_bit(char* bits,unsigned int bit)
> {
>     bits[bit/8] |=3D 1 << (bit&3);
> }
>
> what exactly does the set_bit() function do?
> I don't understand it fully especially the (bit&3) part.
> also I am totally lost in writing a get_bit() function..
>
> any help appreciated!
>
>
> gruss
> /Christian                   mailto:ca...@kr...
>
> ---
>
>
> I propose that the following character sequence for joke markers:
>
> :-)
>
>                              19-Sep-82 11:44    Scott E  Fahlman
>
>
>
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