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 13 5209 ge******@hotmail.com (Elijah Bailey) writes: 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;
}
--
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 p-1;putchar(p[i]\
);}return 0;}
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 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 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 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
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'.
<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.
Ben Pfaff wrote: Elijah Bailey writes:
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?
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
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 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)
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 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
--
Mark Gordon
Paid to be a Geek & a Senior Software Developer
Although my email address says spamtrap, it is real and I read it.
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 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[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;
}
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 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)
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. 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)
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.
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 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.
<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.
In article <bq**********@news-rocq.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 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?
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.
--
Michael
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. 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 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;
}
--
Paul Hsieh http://www.pobox.com/~qed/ http://bstring.sf.net/
Michael Andrew Fyles wrote: In article <bq**********@news-rocq.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 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?
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.
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last post by:
NEW TradeStation 8 Add-on - L'eau Prosper Market Manipulation
Profiling Tools Set By L'eau Prosper Research
Press Release:
L'eau Prosper Research (Website: http://www.leauprosper.com) releases...
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by: Rhadamanthys |
last post by:
Hello All
I am a relative beginner to SQL databases & new to this forum, so please bear with me if my query is too basic and advise if this question belongs somewhere else
I began working at a...
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by: BarryA |
last post by:
What are the essential steps and strategies outlined in the Data Structures and Algorithms (DSA) roadmap for aspiring data scientists? How can individuals effectively utilize this roadmap to progress...
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by: Sonnysonu |
last post by:
This is the data of csv file
1 2 3
1 2 3
1 2 3
1 2 3
2 3
2 3
3
the lengths should be different i have to store the data by column-wise with in the specific length.
suppose the i have to...
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by: Hystou |
last post by:
There are some requirements for setting up RAID:
1. The motherboard and BIOS support RAID configuration.
2. The motherboard has 2 or more available SATA protocol SSD/HDD slots (including MSATA, M.2...
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by: marktang |
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ONU (Optical Network Unit) is one of the key components for providing high-speed Internet services. Its primary function is to act as an endpoint device located at the user's premises. However,...
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by: Hystou |
last post by:
Most computers default to English, but sometimes we require a different language, especially when relocating. Forgot to request a specific language before your computer shipped? No problem! You can...
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by: Oralloy |
last post by:
Hello folks,
I am unable to find appropriate documentation on the type promotion of bit-fields when using the generalised comparison operator "<=>".
The problem is that using the GNU compilers,...
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by: Hystou |
last post by:
Overview:
Windows 11 and 10 have less user interface control over operating system update behaviour than previous versions of Windows. In Windows 11 and 10, there is no way to turn off the Windows...
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by: agi2029 |
last post by:
Let's talk about the concept of autonomous AI software engineers and no-code agents. These AIs are designed to manage the entire lifecycle of a software development project—planning, coding, testing,...
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by: isladogs |
last post by:
The next Access Europe User Group meeting will be on Wednesday 1 May 2024 starting at 18:00 UK time (6PM UTC+1) and finishing by 19:30 (7.30PM).
In this session, we are pleased to welcome a new...
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