P: n/a

I have a long x;
I want to write a function
long f(long x, int k)
such that it extracts every kth 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  
Share this Question
P: n/a
 ge******@hotmail.com (Elijah Bailey) writes: I want to write a function
long f(long x, int k)
such that it extracts every kth 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;
}

int main(void){char p[]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz.\
\n",*q="kl BIcNBFr.NKEzjwCIxNJC";int i=sizeof p/2;char *strchr();int putchar(\
);while(*q){i+=strchr(p,*q++)p;if(i>=(int)sizeof p)i=sizeof p1;putchar(p[i]\
);}return 0;}  
P: n/a

Elijah Bailey wrote: I have a long x; I want to write a function
long f(long x, int k)
such that it extracts every kth 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  
P: n/a

Capstar wrote: Elijah Bailey wrote:
I have a long x; I want to write a function long f(long x, int k)
such that it extracts every kth 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
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 kth 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'.  
P: n/a

<snip> 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.  
P: n/a

Ben Pfaff wrote: Elijah Bailey writes:
I want to write a function
long f(long x, int k)
such that it extracts every kth 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?
Let x_i denote the i_th bit in x counting from right to left.
In particular, x_0 denotes the rightmost bit in x.
Then f(x)_i = x_{k1+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  
P: n/a

On Wed, 03 Dec 2003 10:10:04 +0100
Capstar <sp***@eg.homeip.net> wrote: Capstar wrote: Elijah Bailey wrote:
I have a long x; I want to write a function long f(long x, int k)
such that it extracts every kth 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)
Tut, tut. Assuming CHAR_BIT==8
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 kth 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

Mark Gordon
Paid to be a Geek & a Senior Software Developer
Although my email address says spamtrap, it is real and I read it.  
P: n/a

Elijah Bailey wrote: I have a long x; I want to write a function
long f(long x, int k)
such that it extracts every kth 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 >>= (k1);
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[i];
}
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;
}  
P: n/a

Mark Gordon wrote: On Wed, 03 Dec 2003 10:10:04 +0100 Capstar <sp***@eg.homeip.net> wrote:
Capstar wrote:
Elijah Bailey wrote:
I have a long x; I want to write a function long f(long x, int k)
such that it extracts every kth 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)
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 8bit bytes. I'll keep CHAR_BIT in mind next
time. 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 kth 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.
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 8bit 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.  
P: n/a

On Wed, 03 Dec 2003 17:24:09 +0100
Capstar <sp***@eg.homeip.net> wrote: Mark Gordon wrote:
<snip> Tut, tut. Assuming CHAR_BIT==8
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 8bit 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.
<snip> 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.
It was merely intended as a helpful stylistic comment, not an
instruction.
<snip> No, it's not part of the standard and so when you use ansi pedantic it is not defined by string.h
OK, thanks.
You're welcome.

Mark Gordon
Paid to be a Geek & a Senior Software Developer
Although my email address says spamtrap, it is real and I read it.  
P: n/a

In article <bq**********@newsrocq.inria.fr>, Grumble wrote: Ben Pfaff wrote: Elijah Bailey writes:
I want to write a function
long f(long x, int k)
such that it extracts every kth 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?
Let x_i denote the i_th bit in x counting from right to left.
In particular, x_0 denotes the rightmost bit in x.
Then f(x)_i = x_{k1+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
rightmost bit in x, that would make x_8 one to
the left of the leftmost bit in x. For the
example given, this doesn't exist.

Michael  
P: n/a

Mark Gordon wrote: On Wed, 03 Dec 2003 17:24:09 +0100 Capstar <sp***@eg.homeip.net> wrote:
Mark Gordon wrote:
<snip>
Tut, tut. Assuming CHAR_BIT==8
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.
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.
<snip>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.
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.  
P: n/a
 ge******@hotmail.com says... I have a long x; I want to write a function
long f(long x, int k)
such that it extracts every kth 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 >> (k1)) & 1;
default: break;
}
return 0;
}

Paul Hsieh http://www.pobox.com/~qed/ http://bstring.sf.net/  
P: n/a

Michael Andrew Fyles wrote: In article <bq**********@newsrocq.inria.fr>, Grumble wrote:
Ben Pfaff wrote:
Elijah Bailey writes:
I want to write a function
long f(long x, int k)
such that it extracts every kth 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?
Let x_i denote the i_th bit in x counting from right to left.
In particular, x_0 denotes the rightmost bit in x.
Then f(x)_i = x_{k1+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 rightmost bit in x, that would make x_8 one to the left of the leftmost bit in x. For the example given, this doesn't exist.
You are correct.   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 4903
 replies: 13
 date asked: Nov 13 '05
