ref: 999d97f6bb29799cc9dbaf9499067fe9ad85cd52
dir: /bitvector.c/
/*
bit vector primitives
todo:
* reverse
* nreverse
(- rotate left/right)
* shl_to
* not
- shr_row, shl_row
These routines are the back end supporting bit matrices. Many operations
on bit matrices are slow (such as accessing or setting a single element!)
but certain operations are privileged and lend themselves to extremely
efficient implementation due to the bit-vector nature of machine integers.
These are:
done:
& | $ ~ copy reverse fill sum prod
todo:
shift trans rowswap
would be nice:
channel interleave
Important note:
Out-of-place functions always assume dest and source have the same amount
of space available.
shr_to, shl_to, not_to, and reverse_to assume source and dest don't overlap
and_to, or_to, and xor_to allow overlap.
*/
#include "llt.h"
uint32_t *
bitvector_resize(uint32_t *b, uint64_t oldsz, uint64_t newsz, int initzero)
{
uint32_t *p;
size_t sz = ((newsz+31)>>5) * sizeof(uint32_t);
p = LLT_REALLOC(b, sz);
if(p == nil)
return nil;
if(initzero && newsz>oldsz){
size_t osz = ((oldsz+31)>>5) * sizeof(uint32_t);
memset(&p[osz/sizeof(uint32_t)], 0, sz-osz);
}
return p;
}
uint32_t *
bitvector_new(uint64_t n, int initzero)
{
return bitvector_resize(nil, 0, n, initzero);
}
size_t
bitvector_nwords(uint64_t nbits)
{
return (nbits+31)>>5;
}
void
bitvector_set(uint32_t *b, uint64_t n, uint32_t c)
{
if(c)
b[n>>5] |= 1<<(n&31);
else
b[n>>5] &= ~(1<<(n&31));
}
uint32_t
bitvector_get(uint32_t *b, uint64_t n)
{
return b[n>>5] & (1<<(n&31));
}
static int
ntz(uint32_t x)
{
int n;
if(x == 0)
return 32;
n = 1;
if((x & 0x0000FFFF) == 0){
n = n +16;
x = x >>16;
}
if((x & 0x000000FF) == 0){
n = n + 8;
x = x >> 8;
}
if((x & 0x0000000F) == 0){
n = n + 4;
x = x >> 4;
}
if((x & 0x00000003) == 0){
n = n + 2;
x = x >> 2;
}
return n - (x & 1);
}
// given a bitvector of n bits, starting at bit n0 find the next
// set bit, including n0.
// returns n if no set bits.
uint32_t
bitvector_next(uint32_t *b, uint64_t n0, uint64_t n)
{
if(n0 >= n)
return n;
uint32_t i = n0>>5;
uint32_t nb = n0&31;
uint32_t nw = (n+31)>>5;
uint32_t w;
if(i < nw-1 || (n&31) == 0)
w = b[i]>>nb;
else
w = (b[i]&lomask(n&31)) >> nb;
if(w != 0)
return ntz(w) + n0;
if(i == nw-1)
return n;
i++;
while(i < nw-1){
w = b[i];
if(w != 0)
return ntz(w) + (i<<5);
i++;
}
w = b[i];
nb = n&31;
i = ntz(w);
if(nb == 0)
return i + (n-32);
if(i >= nb)
return n;
return i + (n-nb);
}