using binary numbers in c

Hi everyone:
when we wanna use hex numbers in C, we usually write something like:
int hex_num = 0x12F9;

but how can I declare a binary number in a similar way by putting some
leading words to tell the complier this is a binary number???
similarly, in printf, we have %d for an decimal number %x and %o for hex and
octal numbers... how about binary numbers??

Thank you very much...
--
{ Kelvin@!!! }

On Thu, 06 May 2004 01:18:56 GMT, "Kelvin@!!!" <>
wrote:

>Hi everyone:
>when we wanna use hex numbers in C, we usually write something like:
>int hex_num = 0x12F9;
>
>but how can I declare a binary number in a similar way by putting some
>leading words to tell the complier this is a binary number???
>similarly, in printf, we have %d for an decimal number %x and %o for hex and
>octal numbers... how about binary numbers??
>
>Thank you very much...

Here's something I picked up from, I believe, this very newsgroup not long
back:

------------- macros.h: ---------------------------------

/*
macros.h:
Binary constant generator macro
By Tom Torfs - donated to the public domain
*/

/* All macro's evaluate to compile-time constants */

/* *** helper macros *** */

/* turn a numeric literal into a hex constant
(avoids problems with leading zeroes)
8-bit constants max value 0x11111111, always fits in unsigned long
*/
#define HEX__(n) 0x##n##LU

/* 8-bit conversion function */
#define B8__(x) ((x&0x0000000FLU)?1:0) \
+((x&0x000000F0LU)?2:0) \
+((x&0x00000F00LU)?4:0) \
+((x&0x0000F000LU)?8:0) \
+((x&0x000F0000LU)?16:0) \
+((x&0x00F00000LU)?32:0) \
+((x&0x0F000000LU)?64:0) \
+((x&0xF0000000LU)?128:0)

/* *** user macros *** */

/* for upto 8-bit binary constants */
#define B8(d) ((unsigned char)B8__(HEX__(d)))

/* for upto 16-bit binary constants, MSB first */
#define B16(dmsb,dlsb) (((unsigned short)B8(dmsb)<< \
+ B8(dlsb))

/* for upto 32-bit binary constants, MSB first */
#define B32(dmsb,db2,db3,dlsb) (((unsigned long)B8(dmsb)<<24) \
+ ((unsigned long)B8(db2)<<16) \
+ ((unsigned long)B8(db3)<< \
+ B8(dlsb))

/* Sample usage:
B8(01010101) = 85
B16(10101010,01010101) = 43605
B32(10000000,11111111,10101010,01010101) = 2164238933
*/

------------ test.c ---------------------

#include <stdio.h>
#include "macros.h"

int main()
{
int i = B8(1010);
int j = B8(10000000);

printf("i = %d, j = %d\n", i, j);
return 0;
}


HTH,
-leor


--
Leor Zolman --- BD Software --- www.bdsoft.com
On-Site Training in C/C++, Java, Perl and Unix
C++ users: download BD Software's free STL Error Message Decryptor at:
www.bdsoft.com/tools/stlfilt.html

"Kelvin@!!!" <> wrote in message
news:4ogmc.384222$Ig.25579@pd7tw2no...
> Hi everyone:
> when we wanna use hex numbers in C, we usually write something like:
> int hex_num = 0x12F9;
>
> but how can I declare a binary number in a similar way by putting some
> leading words to tell the complier this is a binary number???

First let me dispel what I think is an erroneous notion you have.
All numbers are stored as 'binary'. When you talk about e.g.
'decimal', 'binary', 'hex', etc., you're talking about the
*textual* representation of a number, e.g. using digits 0-9,
0 and 1, 0 - F, etc.

That said, no, C has no syntax for expressing a binary pattern in source
code. Hex is as close (and imo more 'convenient') as it gets.

> similarly, in printf, we have %d for an decimal number %x and %o for hex
and
> octal numbers... how about binary numbers??

Nope, printf() doesn't have a type specifier for "zero and one"
representation. But it's trivial to write a function
to produce a string of zero and one characters from an integer,
then use %s with its output.

Hints:

x % 2;
x /= 2;

A function could also be written to input a string containing
zeros and ones and convert it to a numeric type.

-Mike

"Leor Zolman" <> wrote in message
news:...
> On Thu, 06 May 2004 01:18:56 GMT, "Kelvin@!!!" <>
> wrote:
>
> >Hi everyone:
> >when we wanna use hex numbers in C, we usually write something like:
> >int hex_num = 0x12F9;
> >
> >but how can I declare a binary number in a similar way by putting some
> >leading words to tell the complier this is a binary number???
> >similarly, in printf, we have %d for an decimal number %x and %o for hex
and
> >octal numbers... how about binary numbers??
> >
> >Thank you very much...
>
> Here's something I picked up from, I believe, this very newsgroup not long
> back:
>
> ------------- macros.h: ---------------------------------
>
> /*
> macros.h:
> Binary constant generator macro
> By Tom Torfs - donated to the public domain
> */
>
> /* All macro's evaluate to compile-time constants */
>
> /* *** helper macros *** */
>
> /* turn a numeric literal into a hex constant
> (avoids problems with leading zeroes)
> 8-bit constants max value 0x11111111, always fits in unsigned long
> */
> #define HEX__(n) 0x##n##LU
>
> /* 8-bit conversion function */
> #define B8__(x) ((x&0x0000000FLU)?1:0) \
> +((x&0x000000F0LU)?2:0) \
> +((x&0x00000F00LU)?4:0) \
> +((x&0x0000F000LU)?8:0) \
> +((x&0x000F0000LU)?16:0) \
> +((x&0x00F00000LU)?32:0) \
> +((x&0x0F000000LU)?64:0) \
> +((x&0xF0000000LU)?128:0)
>
> /* *** user macros *** */
>
> /* for upto 8-bit binary constants */
> #define B8(d) ((unsigned char)B8__(HEX__(d)))
>
> /* for upto 16-bit binary constants, MSB first */
> #define B16(dmsb,dlsb) (((unsigned short)B8(dmsb)<< \
> + B8(dlsb))
>
> /* for upto 32-bit binary constants, MSB first */
> #define B32(dmsb,db2,db3,dlsb) (((unsigned long)B8(dmsb)<<24) \
> + ((unsigned long)B8(db2)<<16) \
> + ((unsigned long)B8(db3)<< \
> + B8(dlsb))
>
> /* Sample usage:
> B8(01010101) = 85
> B16(10101010,01010101) = 43605
> B32(10000000,11111111,10101010,01010101) = 2164238933
> */
>
> ------------ test.c ---------------------
>
> #include <stdio.h>
> #include "macros.h"
>
> int main()
> {
> int i = B8(1010);
> int j = B8(10000000);
>
> printf("i = %d, j = %d\n", i, j);
> return 0;
> }
>
>
> HTH,
> -leor
>
>
> --
> Leor Zolman --- BD Software --- www.bdsoft.com
> On-Site Training in C/C++, Java, Perl and Unix
> C++ users: download BD Software's free STL Error Message Decryptor at:
> www.bdsoft.com/tools/stlfilt.html

thank you for the code...
but ...
> #define HEX__(n) 0x##n##LU

what does the # stands for here???

--
{ Kelvin@!!! }

Kelvin@!!! <> scribbled the following:
> "Leor Zolman" <> wrote in message
> news:...
>> Here's something I picked up from, I believe, this very newsgroup not long
>> back:

(snip)

> thank you for the code...
> but ...
>> #define HEX__(n) 0x##n##LU

> what does the # stands for here???

Assuming you know what #define means and are asking about the ##:
It's a preprocessor operator that "glues" two preprocessing tokens into
one C token. Here it's used twice, gluing three preprocessing tokens
together. The first is 0x, the second is whatever n gets replaced with,
and the third is LU. For example HEX__(0) would be expanded to a glued
together token 0x0LU. The C compiler itself treats this as a single
token and not as three tokens after each other.

--
/-- Joona Palaste () ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"It sure is cool having money and chicks."
- Beavis and Butt-head

"Kelvin@!!!" <> wrote in message
news:4ogmc.384222$Ig.25579@pd7tw2no...
> Hi everyone:
> when we wanna use hex numbers in C, we usually write something like:
> int hex_num = 0x12F9;
>
> but how can I declare a binary number in a similar way by putting some
> leading words to tell the complier this is a binary number???
> similarly, in printf, we have %d for an decimal number %x and %o for hex
and
> octal numbers... how about binary numbers??
>

There isn't a way to do that, as the compiler does know from binary.

OTOH, you could write a pre-processor that does that, if you wish. It would
take your new symbol, say 0b11110000 and turn it into 0xF0, then pass it
into the compiler.

For printf(), you'd have to write a function to change hex to a binary
string, I surmise. It would have to be at run-time since the variables, are.

--
Mabden

http://www.eskimo.com/~scs/C-faq/q20.11.html

--


~Kieran Simkin
Digital Crocus
http://digital-crocus.com/

"Kelvin@!!!" <> wrote in message
news:4ogmc.384222$Ig.25579@pd7tw2no...
> Hi everyone:
> when we wanna use hex numbers in C, we usually write something like:
> int hex_num = 0x12F9;
>
> but how can I declare a binary number in a similar way by putting some
> leading words to tell the complier this is a binary number???
> similarly, in printf, we have %d for an decimal number %x and %o for hex
and
> octal numbers... how about binary numbers??
>
> Thank you very much...
> --
> { Kelvin@!!! }
>
>

[zuger]

Thank you for the nice example. I changed it slightly to work for my needs in an embedded processor. I type casted with (unsigned long). It seems to work. Why was the 16B type cased differently than the 32B in the original example?


#define HEX__(n) 0x##n##LU

/* 8-bit conversion function */
#define B8__(x) ((x&0x0000000FLU)?1:0) \
+((x&0x000000F0LU)?2:0) \
+((x&0x00000F00LU)?4:0) \
+((x&0x0000F000LU)?8:0) \
+((x&0x000F0000LU)?16:0) \
+((x&0x00F00000LU)?32:0) \
+((x&0x0F000000LU)?64:0) \
+((x&0xF0000000LU)?128:0)

/* *** user macros *** */

/* for upto 8-bit binary constants */
#define B8(d) ((unsigned char)B8__(HEX__(d)))

/* for upto 16-bit binary constants, MSB first */
#define B16(dmsb,dlsb) (((unsigned long)B8(dmsb)<< 8) + (unsigned long)B8(dlsb))


/* for upto 32-bit binary constants, MSB first */
#define B32(dmsb,db2,db3,dlsb) (((unsigned long)B8(dmsb)<<24) \
+ ((unsigned long)B8(db2)<<16) \
+ ((unsigned long)B8(db3)<< 8) \
+ B8(dlsb))

/* Sample usage:
B8(01010101) = 85
B16(10101010,01010101) = 43605
B32(10000000,11111111,10101010,01010101) = 2164238933
*/