bitsperlong
#define longsperbits(n) howmany(n, bitsperlong)
return (bmp->maskp[n/bitsperlong] >> (n % bitsperlong)) & 1;
bmp->maskp[n/bitsperlong] |= 1UL << (n % bitsperlong);
bmp->maskp[n/bitsperlong] &=
~(1UL << (n % bitsperlong));