bit shifts across array elements

[fermineutron]

Lets say i have array

unsigned long X[4];

Now, i want to shift right bits in the array by 5. that is the lowest 5
bits of element N will become the highest 5 bits of element N-1. the
lowest 5 bits of 0th element are lost.

what is the best way to do this?

My C reference book does not go in detail on preserving the bits which
are lost during bitshifts.

X[3]=X[3]>>5;
but what is X[2] in this case?

Also is there a way to determine the count of the most significant
non-zero bit in a variable?
for example in the case

0010010101010010

answer would be 14. This can be done by repeatedly testing the variable
storing above value against 2^N untill 2^N is greater than X. The count
of most significant bit would be N-1, assuming the right ost bit is in
0th position, but is there a more efficient way to do this?

In both cases timing is critical.

Thanks ahead.

 

[fermineutron]

Lets say i have array

unsigned long X[4];

Now, i want to shift right bits in the array by 5. that is the lowest 5
bits of element N will become the highest 5 bits of element N-1. the
lowest 5 bits of 0th element are lost.

what is the best way to do this?

My C reference book does not go in detail on preserving the bits which
are lost during bitshifts.

X[3]=X[3]>>5;
but what is X[2] in this case?

Also is there a way to determine the count of the most significant
non-zero bit in a variable?
for example in the case

0010010101010010

answer would be 14. This can be done by repeatedly testing the variable
storing above value against 2^N untill 2^N is greater than X. The count
of most significant bit would be N-1, assuming the right ost bit is in
0th position, but is there a more efficient way to do this?

In both cases timing is critical.

Thanks ahead.

I guess i could do the following to determine most significan non-zero
bit count:
T stores the value to be tested.
n=0;
while(T>0){
T=T>>1;
n++;
}

is there a beter way?

 

[Chris Johnson]

Lets say i have array

unsigned long X[4];

Now, i want to shift right bits in the array by 5. that is the lowest 5
bits of element N will become the highest 5 bits of element N-1. the
lowest 5 bits of 0th element are lost.

what is the best way to do this?

My C reference book does not go in detail on preserving the bits which
are lost during bitshifts.

X[3]=X[3]>>5;
but what is X[2] in this case?

How about:

X[0] >>= 5;
int i;
for(i = 1; i < arraylen; i++){
/* Move the least significant bits of X to the upper bits of
X[i-1] */
X[i-1] |= X << (8*sizeof(unsigned long) - 5);
X >>=5;
}

Also is there a way to determine the count of the most significant
non-zero bit in a variable?
for example in the case

0010010101010010

answer would be 14. This can be done by repeatedly testing the variable
storing above value against 2^N untill 2^N is greater than X. The count
of most significant bit would be N-1, assuming the right ost bit is in
0th position, but is there a more efficient way to do this?

I use the following:

for(i = 0;n >> i; i++);

i is now the position of the most significant bit (assuming the far
right is bit 1)

 

[Eric Sosman]

Lets say i have array

unsigned long X[4];

Now, i want to shift right bits in the array by 5. that is the lowest 5
bits of element N will become the highest 5 bits of element N-1. the
lowest 5 bits of 0th element are lost.

what is the best way to do this?

My C reference book does not go in detail on preserving the bits which
are lost during bitshifts.

X[3]=X[3]>>5;
but what is X[2] in this case?

Also is there a way to determine the count of the most significant
non-zero bit in a variable?
for example in the case

0010010101010010

answer would be 14. This can be done by repeatedly testing the variable
storing above value against 2^N untill 2^N is greater than X. The count
of most significant bit would be N-1, assuming the right ost bit is in
0th position, but is there a more efficient way to do this?

In both cases timing is critical.

Thanks ahead.

I guess i could do the following to determine most significan non-zero
bit count:
T stores the value to be tested.
n=0;
while(T>0){
T=T>>1;
n++;
}

is there a beter way?

Maybe. Timings are highly machine-dependent, and a technique
that whizzes on one system may wheeze on another. A few ideas:

1) If you know the number of bits in a T-type value you can
do a binary search. For example, if T is a sixteen-bit unsigned
integer you could do

int n = 0;
if (T > 0x00FF) { n = 8; T >>= 8; }
if (T > 0x000F) { n += 4; T >>= 4; }
if (T > 0x0003) { n += 2; T >>= 2; }
n += T >> 1;

1a) Even if you don't know the number of bits but do know a
lower bound, you can use a "big bite" linear search followed by
a binary search as above. For example, if T is an unsigned integer
known to be at least sixteen bits wide but possibly wider,

int n = 0;
while (T > 0xFFFF) { n += 16; T >>= 16; }
/* ... followed by binary search as above */

2) Use a "big bite" linear search to reduce T to a convenient
range and then index a precomputed table with the reduced T.

3) A trick mentioned on this forum within the past few weeks:
Convert T to a floating-point type and extract the exponent.

Before spending much time on these or any other alternatives,
be sure you have solid *evidence* for the criticality of the
timing. Suspicion is not enough.

 

[fermineutron]

Thanks everyone,
all replies were very helpfull.

 

[Samuel Stearley]

Also is there a way to determine the count of the most significant

non-zero bit in a variable?

http://graphics.stanford.edu/~seander/bithacks.html#IntegerLogDeBruijn

Lets say i have array

unsigned long X[4];

Now, i want to shift right bits in the array by 5. that is the lowest 5
bits of element N will become the highest 5 bits of element N-1. the
lowest 5 bits of 0th element are lost.

what is the best way to do this?

My C reference book does not go in detail on preserving the bits which
are lost during bitshifts.

X[3]=X[3]>>5;
but what is X[2] in this case?

Also is there a way to determine the count of the most significant
non-zero bit in a variable?
for example in the case

0010010101010010

answer would be 14. This can be done by repeatedly testing the variable
storing above value against 2^N untill 2^N is greater than X. The count
of most significant bit would be N-1, assuming the right ost bit is in
0th position, but is there a more efficient way to do this?

In both cases timing is critical.

Thanks ahead.

 

[Peter Nilsson]

How about:

X[0] >>= 5;
int i;
for(i = 1; i < arraylen; i++){
/* Move the least significant bits of X to the upper bits
of X[i-1] */
X[i-1] |= X << (8*sizeof(unsigned long) - 5);
X >>=5;
}

The restriction of unpadded integers and 8 bit bytes only is
unnecessary...

X[0] >>= 5;
for(i = 1; i < arraylen; i++)
{
X[i-1] |= X * ((-1ul >> 5) + 1);
X >>=5;
}

[Many compilers will optimise the * to a shift, so you gain portability
without losing efficiency.]