bit manipulation question

[Elijah Bailey]

I have a long x;
I want to write a function

long f(long x, int k)

such that it extracts every k-th bit of x, concatenates
them and returns it. Anyone can help me in writing this
function?

examples
x = 10101010 k = 1 f(x) = 10101010
x = 10101010 k = 2 f(x) = 1111
x = 10101010 k = 3 f(x) = 010
x = 10101010 k = 4 f(x) = 11

Any bit gurus here who can help me?

Thanks,
--Elijah

 

[Ben Pfaff]

I want to write a function

long f(long x, int k)

such that it extracts every k-th bit of x, concatenates
them and returns it. Anyone can help me in writing this
function?

examples
x = 10101010 k = 1 f(x) = 10101010
x = 10101010 k = 2 f(x) = 1111
x = 10101010 k = 3 f(x) = 010
x = 10101010 k = 4 f(x) = 11

I don't understand your k = 3 example. Shouldn't f(x) = 101 for
consistency with the others?

/* I think this'll do the trick, but I haven't tested it or even
compiled it. In fact, I haven't even tried it out on paper,
so you'd better think it through carefully before assuming
anything. */
unsigned long f(unsigned long x, int k)
{
unsigned long out = 0; /* Output so far. */
unsigned long left = -1; /* Mask of bits left to extract from x. */
unsigned long msb = ~(left << 1 >> 1); /* Top bit in a unsigned long. */

do {
/* Extract MSB of x into LSB of out. */
out <<= 1;
if (x & msb)
out |= 1;

/* Advance. */
x <<= k;
left <<= k;
} while (left != 0);

return out;
}

 

[Capstar]

I have a long x;
I want to write a function

long f(long x, int k)

such that it extracts every k-th bit of x, concatenates
them and returns it. Anyone can help me in writing this
function?

examples
x = 10101010 k = 1 f(x) = 10101010
x = 10101010 k = 2 f(x) = 1111
x = 10101010 k = 3 f(x) = 010
x = 10101010 k = 4 f(x) = 11

Any bit gurus here who can help me?

Thanks,
--Elijah

Although this seems to me like a homework asignment, I couldn't resist
to try something.
I wouldn't try calling it as you do in your example, because I don't
think any C compiler would compile it.

This code is untested and does not take care any special cases. There is
at least on obvious one, but you can take care of that yourself.

long f(long x, int k)
{
long temp = 0;
long i, j;

for(i = 0, j = 0; i < sizeof(x) * 8; i += k, ++j)
temp |= (x & (1 << i)) >> (i - j);

return temp;
}

Mark

 

[Capstar]

Although this seems to me like a homework asignment, I couldn't resist
to try something.
I wouldn't try calling it as you do in your example, because I don't
think any C compiler would compile it.

This code is untested and does not take care any special cases. There is
at least on obvious one, but you can take care of that yourself.

long f(long x, int k)
{
long temp = 0;
long i, j;

for(i = 0, j = 0; i < sizeof(x) * 8; i += k, ++j)
temp |= (x & (1 << i)) >> (i - j);

return temp;
}

Mark

Hmmm, after reading your post again, I seem to have made a little
mistake. In my previous post you always start at bit 0. But when you say
take every k-th bit, you obviously want to start at bit k - 1.

I also tested it so I'll post my complete test program:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char *bitprint(long x, unsigned long max)
{
unsigned long i, j;
static char bitstring[sizeof(x) * 8 + 1];

if(max > sizeof(x) * 8) max = sizeof(x) * 8;

for(i = max - 1, j = 0; j < max; --i, ++j)
{
if(x & (1 << i)) bitstring[j] = '1';
else bitstring[j] = '0';
}
bitstring[max] = '\0';

return bitstring;
}

long f(long x, int k)
{
long temp = 0;
unsigned long i, j;

for(i = k - 1, j = 0; i < sizeof(x) * 8; i += k, ++j)
temp |= (x & (1 << i)) >> (i - j);

return temp;
}

int main(void)
{
long x = 0xaa; /*10101010*/
int k;
char *xstring;

xstring = strdup(bitprint(x, 8));

for(k = 1; k <= 4; ++k)
printf("x = %s\tk = %d\t f(x, k) = %s\n", xstring, k, bitprint(f(x,
k), 8));

free(xstring);

return 0;
}

Mark

PS. Isn't strdup in the standard? If I compile this with gcc -W -Wall
-ansi -pedantic I get: warning: implicit declaration of function `strdup'.

 

[nrk]

PS. Isn't strdup in the standard? If I compile this with gcc -W -Wall
-ansi -pedantic I get: warning: implicit declaration of function `strdup'.

No, strdup is *not* part of the standard.

<OT>
According to my manpage, conforms to SYS V R3 and BSD 4.3
</OT>

-nrk.

 

[Mark Gordon]

Tut, tut. Assuming CHAR_BIT==8

Hmmm, after reading your post again, I seem to have made a little
mistake. In my previous post you always start at bit 0. But when you
say take every k-th bit, you obviously want to start at bit k - 1.

I also tested it so I'll post my complete test program:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char *bitprint(long x, unsigned long max)

I would prefer to have the function take a parameter to the buffer
to be used rather than returning a pointer to a static buffer.

{
unsigned long i, j;
static char bitstring[sizeof(x) * 8 + 1];

Tut, tut. Still assuming CHAR_BIT==8 throughout this new version.

if(max > sizeof(x) * 8) max = sizeof(x) * 8;

for(i = max - 1, j = 0; j < max; --i, ++j)
{
if(x & (1 << i)) bitstring[j] = '1';
else bitstring[j] = '0';
}
bitstring[max] = '\0';

return bitstring;
}

long f(long x, int k)
{
long temp = 0;
unsigned long i, j;

for(i = k - 1, j = 0; i < sizeof(x) * 8; i += k, ++j)
temp |= (x & (1 << i)) >> (i - j);

return temp;
}

int main(void)
{
long x = 0xaa; /*10101010*/
int k;
char *xstring;

xstring = strdup(bitprint(x, 8));

for(k = 1; k <= 4; ++k)
printf("x = %s\tk = %d\t f(x, k) = %s\n", xstring, k,
bitprint(f(x,
k), 8));

free(xstring);

return 0;
}

Mark

PS. Isn't strdup in the standard? If I compile this with gcc -W -Wall
-ansi -pedantic I get: warning: implicit declaration of function
`strdup'.

No, it's not part of the standard and so when you use -ansi -pedantic it
is not defined by string.h

 

[Grumble]

I don't understand your k = 3 example. Shouldn't f(x) = 101 for
consistency with the others?

Let x_i denote the i_th bit in x counting from right to left.

In particular, x_0 denotes the right-most bit in x.

Then f(x)_i = x_{k-1+i*k}

k=1 => f(x)_i = x_i
k=2 => f(x)_i = x_{2*i+1}
k=3 => f(x)_i = x_{3*i+2}

Thus, from right to left, x_2 | x_5 | x_8

 

[nrk]

I have a long x;
I want to write a function

long f(long x, int k)

such that it extracts every k-th bit of x, concatenates
them and returns it. Anyone can help me in writing this
function?

examples
x = 10101010 k = 1 f(x) = 10101010
x = 10101010 k = 2 f(x) = 1111
x = 10101010 k = 3 f(x) = 010
x = 10101010 k = 4 f(x) = 11

Any bit gurus here who can help me?

Thanks,
--Elijah

Probably a bit of beating around the bush going on in the code below, but
hey, you can verify your function as well

-nrk.

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

unsigned long bitextract(unsigned long input, int k) {
unsigned long ret = 0;
size_t i;
int j;

if ( k <= 1 ) return input;

for ( i = 0, j = 0; i < sizeof input * CHAR_BIT; i += k, ++j ) {
input >>= (k-1);
ret |= (input & 1) << j;
input >>= 1;
}

return ret;
}

char *printBinary(char *buf, unsigned long num) {
static char *hex2bin[] =
{
"0000", "0001", "0010", "0011",
"0100", "0101", "0110", "0111",
"1000", "1001", "1010", "1011",
"1100", "1101", "1110", "1111"
};

size_t i;

for ( i = 4; i <= (sizeof num * CHAR_BIT); i += 4 ) {
int shift = (sizeof num * CHAR_BIT) - i;
memcpy(buf, hex2bin[((num & (0xFUL << shift)) >> shift)], 4);
buf += 4;
}
*buf = 0;

return buf - (i - 4);
}

int verifyResult(int input, int k, int result) {
char input_str[sizeof input * CHAR_BIT + 1];
char result_str[sizeof input * CHAR_BIT + 1];
char extract_str[sizeof input * CHAR_BIT + 1];
int i, j;

printBinary(input_str, input);
printBinary(result_str, result);

for ( i = sizeof input_str - 1 - k, j = sizeof extract_str - 2;
i >= 0;
i -= k, --j )
{
extract_str[j] = input_str;
}

while ( j >= 0 ) extract_str[j--] = '0';

extract_str[sizeof extract_str - 1] = 0;

return strcmp(result_str, extract_str);
}

int main(void) {
unsigned long input = 0x12345678UL;
unsigned long result;
size_t i;

for ( i = 1; i <= sizeof input * CHAR_BIT; ++i ) {
result = bitextract(input, i);
if ( verifyResult(input, i, result) ) {
printf("Failed for bit %d\n", i);
break;
}
}

if ( i >= sizeof input * CHAR_BIT )
printf("Success!\n");

return 0;
}

 

[Capstar]

Tut, tut. Assuming CHAR_BIT==8

I am not a c.l.c. guru and so I'm not an allmighty know it all. And I
thought sizeof returned in 8-bit bytes. I'll keep CHAR_BIT in mind next
time.

I would prefer to have the function take a parameter to the buffer
to be used rather than returning a pointer to a static buffer.

I don't realy care wat you want. I just needed a function, which would
return a character array so I could display the value of f(x,k) to
verify the results easily.

{
unsigned long i, j;
static char bitstring[sizeof(x) * 8 + 1];

Tut, tut. Still assuming CHAR_BIT==8 throughout this new version.

I didn't became an allmighty know it all c.l.c. guru in those 32
minutes, So I still thought sizeof returned in 8-bit bytes.

if(max > sizeof(x) * 8) max = sizeof(x) * 8;

for(i = max - 1, j = 0; j < max; --i, ++j)
{
if(x & (1 << i)) bitstring[j] = '1';
else bitstring[j] = '0';
}
bitstring[max] = '\0';

return bitstring;
}

long f(long x, int k)
{
long temp = 0;
unsigned long i, j;

for(i = k - 1, j = 0; i < sizeof(x) * 8; i += k, ++j)
temp |= (x & (1 << i)) >> (i - j);

return temp;
}

int main(void)
{
long x = 0xaa; /*10101010*/
int k;
char *xstring;

xstring = strdup(bitprint(x, 8));

for(k = 1; k <= 4; ++k)
printf("x = %s\tk = %d\t f(x, k) = %s\n", xstring, k,
bitprint(f(x,
k), 8));

free(xstring);

return 0;
}

Mark

PS. Isn't strdup in the standard? If I compile this with gcc -W -Wall
-ansi -pedantic I get: warning: implicit declaration of function
`strdup'.

No, it's not part of the standard and so when you use -ansi -pedantic it
is not defined by string.h

OK, thanks.

 

[Mark Gordon]

Mark Gordon wrote:


I am not a c.l.c. guru and so I'm not an allmighty know it all.

Nor am I. Perhaps I should not have said "Tut, tut" since it was not my
intent to cause offence.

And I
thought sizeof returned in 8-bit bytes. I'll keep CHAR_BIT in mind
next time.

Well, I've been caught out a few times here. One of the good things
about this group is that people do tend to pick up on the assumptions we
(and I do deliberately include myself) make.

I don't realy care wat you want. I just needed a function, which would
return a character array so I could display the value of f(x,k) to
verify the results easily.

It was merely intended as a helpful stylistic comment, not an
instruction.

OK, thanks.

You're welcome.

 

[Michael Andrew Fyles]

Let x_i denote the i_th bit in x counting from right to left.

In particular, x_0 denotes the right-most bit in x.

Then f(x)_i = x_{k-1+i*k}

k=1 => f(x)_i = x_i
k=2 => f(x)_i = x_{2*i+1}
k=3 => f(x)_i = x_{3*i+2}

Thus, from right to left, x_2 | x_5 | x_8

Unless I've misunderstood, if x_0 is the
right-most bit in x, that would make x_8 one to
the left of the left-most bit in x. For the
example given, this doesn't exist.

 

[Capstar]

Nor am I. Perhaps I should not have said "Tut, tut" since it was not my
intent to cause offence.

OK, well then no offence taken. It was indeed the "Tut, tut" that pissed
me off. I normally don't mind any comment on my code, I even welcome it.

It was merely intended as a helpful stylistic comment, not an
instruction.

I know. Sorry, I normally wouldn't react like that. (Was still pissed).
But your solution indeed has it's advantages. Mine isn't useable more
than once in a printf statement for instance. That's why I did:

xstring = strdup(bitprint(x, 8));

Mark.

 

[Paul Hsieh]

(e-mail address removed) says...

I have a long x;
I want to write a function

long f(long x, int k)

such that it extracts every k-th bit of x, concatenates
them and returns it. Anyone can help me in writing this
function?

examples
x = 10101010 k = 1 f(x) = 10101010
x = 10101010 k = 2 f(x) = 1111
x = 10101010 k = 3 f(x) = 010
x = 10101010 k = 4 f(x) = 11

Any bit gurus here who can help me?

Its not portable, but I'm sure you can adapt it to your platform of choice:

long f(long x, int k) {
switch (k) {
case 1: return x;
case 2: x &= 0xAAAAAAAA;
x = (x | (x >> 1)) & 0x66666666;
x = (x | (x >> 2)) & 0x1E1E1E1E;
x = (x | (x >> 4)) & 0x01FE01FE;
x = (x | (x >> 8)) & 0x0001FFFE;
return x >> 1;
case 3: x &= 0x24924924;
x = (x | (x >> 2)) & 0x0C30C30C;
x = (x | (x >> 4)) & 0x0C03C03C;
x = (x | (x >> 8)) & 0x0C0003FC;
x = (x | (x >> 16)) & 0x00000FFC;
return x >> 2;
case 4: x &= 0x88888888;
x = (x | (x >> 3)) & 0x18181818;
x = (x | (x >> 6)) & 0x00780078;
x = (x | (x >> 12)) & 0x000007F8;
return x >> 3;
case 5: x &= 0x21084210;
x = (x | (x >> 4)) & 0x0300C030;
x = (x | (x >> 8)) & 0x030000F0;
x = (x | (x >> 16)) & 0x000003F0;
return x >> 4;
case 6: x &= 0x20820820;
x = (x | (x >> 5)) & 0x20060060;
x = (x | (x >> 10)) & 0x200001E0;
x = (x | (x >> 20)) & 0x000003E0;
return x >> 5;
case 7: x &= 0x08102040;
x = (x | (x >> 6)) & 0x003000C0;
x = (x | (x >> 12)) & 0x000003C0;
return x >> 6;
case 8: x &= 0x80808080;
x = (x | (x >> 7)) & 0x01800180;
x = (x | (x >> 14)) & 0x00000780;
return x >> 7;
case 9: x &= 0x04020100;
x = (x | (x >> 8)) & 0x04000300;
x = (x | (x >> 16)) & 0x00000700;
return x >> 8;
case 10: x &= 0x20080200;
x = (x | (x >> 9)) & 0x20000600;
x = (x | (x >> 18)) & 0x00000E00;
return x >> 9;
case 11: x &= 0x00200400;
x = (x | (x >> 10)) & 0x00000C00;
return x >> 10;
case 12: x &= 0x00800800;
x = (x | (x >> 11)) & 0x00001800;
return x >> 11;
case 13: x &= 0x02001000;
x = (x | (x >> 12)) & 0x00003000;
return x >> 12;
case 14: x &= 0x08002000;
x = (x | (x >> 13)) & 0x00006000;
return x >> 13;
case 15: x &= 0x20004000;
x = (x | (x >> 14)) & 0x0000C000;
return x >> 14;
case 16: x &= 0x80008000;
x = (x | (x >> 15)) & 0x00018000;
return x >> 15;
case 17: case 18: case 19: case 20: case 21:
case 22: case 23: case 24: case 25: case 26:
case 27: case 28: case 29: case 30: case 31:
case 32: return (x >> (k-1)) & 1;
default: break;
}
return 0;
}

 

[Grumble]

Unless I've misunderstood, if x_0 is the
right-most bit in x, that would make x_8 one to
the left of the left-most bit in x. For the
example given, this doesn't exist.

You are correct.