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using binary numbers in c

 
 
Kelvin@!!!
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Posts: n/a
 
      05-06-2004
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@!!! }


 
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Leor Zolman
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Posts: n/a
 
      05-06-2004
On Thu, 06 May 2004 01:18:56 GMT, "Kelvin@!!!" <(E-Mail Removed)>
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
 
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Mike Wahler
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      05-06-2004

"Kelvin@!!!" <(E-Mail Removed)> 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


 
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Kelvin@!!!
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Posts: n/a
 
      05-06-2004

"Leor Zolman" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> On Thu, 06 May 2004 01:18:56 GMT, "Kelvin@!!!" <(E-Mail Removed)>
> 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@!!! }


 
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Joona I Palaste
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Posts: n/a
 
      05-06-2004
Kelvin@!!! <(E-Mail Removed)> scribbled the following:
> "Leor Zolman" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
>> 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 ((E-Mail Removed)) ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"It sure is cool having money and chicks."
- Beavis and Butt-head
 
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Mabden
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Posts: n/a
 
      05-06-2004
"Kelvin@!!!" <(E-Mail Removed)> 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


 
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Kieran Simkin
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Posts: n/a
 
      05-08-2004
http://www.eskimo.com/~scs/C-faq/q20.11.html

--


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

"Kelvin@!!!" <(E-Mail Removed)> 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@!!! }
>
>



 
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zuger zuger is offline
Junior Member
Join Date: Jul 2009
Posts: 1
 
      07-25-2009
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
*/
 
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cchris cchris is offline
Junior Member
Join Date: Aug 2010
Posts: 1
 
      08-06-2010
Thanks for this.
The code above inspired me to a version that does _not_ compile if non binary values are entered:

typedef enum
{
b0000 = 0x00,
b0001 = 0x01,
b0010 = 0x02,
b0011 = 0x03,
b0100 = 0x04,
b0101 = 0x05,
b0110 = 0x06,
b0111 = 0x07,
b1000 = 0x08,
b1001 = 0x09,
b1010 = 0x0A,
b1011 = 0x0B,
b1100 = 0x0C,
b1101 = 0x0D,
b1110 = 0x0E,
b1111 = 0x0F
} T_BIN_NIBBLE;

/* 2 nibbles to 8 bit */
#define B8(msn,lsn) ( ( b##msn<<4 ) \
|( b##lsn ) )

/* 4 nibbles to 16 bit */
#define B16(msn,n2, n1,lsn) ( (unsigned short)( b##msn<<12 ) \
|(unsigned short)( b##n2 <<8 ) \
|(unsigned short)( b##n1 <<4 ) \
|(unsigned short)( b##lsn ) )

/* Sample usage:
B8(0101,0101) = 85
B16(1010,1010 , 0101,0101) = 43605
*/
 
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rchandelier rchandelier is offline
Junior Member
Join Date: Sep 2011
Posts: 1
 
      09-28-2011
A few compilers (usually Microcontroller's ones) has a special feature implemented within recognizing literal binary numbers by prefix "0b..." preceding the number, although most compilers (C/C++ standards) don't have such feature and if it is the case, here it is my alternative solution:

#define B_0000 0
#define B_0001 1
#define B_0010 2
#define B_0011 3
#define B_0100 4
#define B_0101 5
#define B_0110 6
#define B_0111 7
#define B_1000 8
#define B_1001 9
#define B_1010 a
#define B_1011 b
#define B_1100 c
#define B_1101 d
#define B_1110 e
#define B_1111 f

#define _B2H(bits) B_##bits
#define B2H(bits) _B2H(bits)
#define _HEX(n) 0x##n##UL
#define HEX(n) _HEX(n)
#define _CCAT(a,b) a##b
#define CCAT(a,b) _CCAT(a,b)

#define BYTE(a,b) HEX( CCAT(B2H( a),B2H( b)) )
#define WORD(a,b,c,d) HEX( CCAT(CCAT(B2H( a),B2H( b)),CCAT(B2H( c),B2H( d))) )
#define DWORD(a,b,c,d,e,f,g,h) HEX( CCAT( CCAT(CCAT(B2H( a),B2H( b)),CCAT(B2H( c),B2H( d))),CCAT(CCAT(B2H( e),B2H( f)),CCAT(B2H( g),B2H( h))) ) )

//using example
char b = BYTE(0100,0001); //equivalent to b = 65; or b = 'A'; or b = 0x41;
unsigned int w = WORD(1101,1111,0100,0011); //equivalent to w = 57155; or w = 0xdf43;
unsigned long int dw = DWORD(1101,1111,0100,0011,1111,1101,0010,1000); //equivalent to dw = 3745774888; or dw = 0xdf43fd28;

Disadvantages: (it's not such a big ones)
- The binary numbers have to be grouped 4 by 4;
- The binary literals have to be only unsigned integer numbers;

Advantages:
- Total preprocessor driven, not spending processor time in pointless operations (like "?.. :..", "<<", "+") to the executable program (it may be performed hundred of times in the final application);
- It works "mainly in C" compilers and C++ as well (template+enum solution works only in C++ compilers);
- It has only the limitation of "longness" for expressing "literal constant" values. There would have been earlyish longness limitation (usually 8bits:0-255) if one had expressed constant values by parsing resolve of "enum solution"(usually 255 = reach enum definition limit), differently, "literal constant" limitations, in the compiler allows greater numbers;
- Some other solutions demand exagerated number of constant definitions (#define's in my opinion) including long or several header files (in most cases not easily readable and understandable, and make the project become unnecessarily confused and extended, like that using "BOOST_BINARY()");
- Simplicity of the solution: easily readable, understandable and adjustable for other cases (could be extended for grouping 8 by 8 too);

I hope it helps, thanks. Renato Chandelier.

 
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