Help on bit operation

Help on bit operation

[Randi Botse]

Hi, I'm beginner C programmer, i have a problem, i want to store some
information in a integer, a integer will be 32 bit on my machine, i
want to have as follow:

6 bit (information 1)   MSB
4 bit (information 2)
8 bit (information 3)
5 bit (information 4)
9 bit (information 5)   LSB

For example i set the informations as follow (in decimal):

information 1 = 43        or 101011
information 2 = 11        or 1011
information 3 = 120      or 1111000
information 3 = 30        or 11110
information 4 = 418      or 110100010

if i join all informations i should get a 32 bit integer valued
2935782212 or 01010111011111100011110110100010, then my problem is how
to retrieve these informations on bit operation? i want to know what's
the value of information-2 or information-3, etc directly.  And, is
there any good way to join these informations to be an 32 bit integer?

at this time i convert the 32bit integer into binary string, process
it's with array segment to get all informations then convert them to
integer,
to build the 32bit integer, i join all information value into binary
string (yes, 32 bit of char ;p) join all of them then convert to
integer.

I know my way is sucks and too far away from COOL thing ;p,  i think
there are cool way to do this!.

Thanks before!
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Re: Help on bit operation

[Akos Marton]

Hello Randi,

I suppose for you the bitfields is useful:
struct {
  int information_1: 6;
  int information_2: 4;
  int information_3: 8;
  int information_4: 5;
  int information_5: 9;
};
This solution is reduce the usage of memory, but needs more
computation in general C programming.
May there another solution using 'union {};' ?

mAkos

On Thu, Aug 27, 2009 at 1:02 AM, Randi Botse<nightdecoder@gmail.com> wrote:
> Hi, I'm beginner C programmer, i have a problem, i want to store some
> information in a integer, a integer will be 32 bit on my machine, i
> want to have as follow:
>
> 6 bit (information 1)   MSB
> 4 bit (information 2)
> 8 bit (information 3)
> 5 bit (information 4)
> 9 bit (information 5)   LSB
>
> For example i set the informations as follow (in decimal):
>
> information 1 = 43        or 101011
> information 2 = 11        or 1011
> information 3 = 120      or 1111000
> information 3 = 30        or 11110
> information 4 = 418      or 110100010
>
> if i join all informations i should get a 32 bit integer valued
> 2935782212 or 01010111011111100011110110100010, then my problem is how
> to retrieve these informations on bit operation? i want to know what's
> the value of information-2 or information-3, etc directly.  And, is
> there any good way to join these informations to be an 32 bit integer?
>
> at this time i convert the 32bit integer into binary string, process
> it's with array segment to get all informations then convert them to
> integer,
> to build the 32bit integer, i join all information value into binary
> string (yes, 32 bit of char ;p) join all of them then convert to
> integer.
>
> I know my way is sucks and too far away from COOL thing ;p,  i think
> there are cool way to do this!.
>
> Thanks before!
> --
> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
> the body of a message to majordomo@vger.kernel.org
> More majordomo info at  http://vger.kernel.org/majordomo-info.html
>



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Re: Help on bit operation

[Uday Verma]

If the sizes of your infos is fixed you can do something like this:

info5 = packet & 0x1FF;
info4 = (packet >> 9) & 0x1F;
.. and so on

note that I am masking out unwanted parts of the 32 bit packet using
an all 1s  bit pattern 0x1FF : 1 1111 1111, 0x1F : 1 1111.  in case of
info4, I first need to move the value 9 bits to the right so that 5
bits of info4 form the LSB.

You can form the packet similarly,

unsigned int packet = ((info1 & 0x3F) << 26) | ((info2 & 0xF) << 22) ...

HTH

packet is your single 32-bit integer which has all these infos encoded.

On Wed, Aug 26, 2009 at 6:02 PM, Randi Botse<nightdecoder@gmail.com> wrote:
> Hi, I'm beginner C programmer, i have a problem, i want to store some
> information in a integer, a integer will be 32 bit on my machine, i
> want to have as follow:
>
> 6 bit (information 1)   MSB
> 4 bit (information 2)
> 8 bit (information 3)
> 5 bit (information 4)
> 9 bit (information 5)   LSB
>
> For example i set the informations as follow (in decimal):
>
> information 1 = 43        or 101011
> information 2 = 11        or 1011
> information 3 = 120      or 1111000
> information 3 = 30        or 11110
> information 4 = 418      or 110100010
>
> if i join all informations i should get a 32 bit integer valued
> 2935782212 or 01010111011111100011110110100010, then my problem is how
> to retrieve these informations on bit operation? i want to know what's
> the value of information-2 or information-3, etc directly.  And, is
> there any good way to join these informations to be an 32 bit integer?
>
> at this time i convert the 32bit integer into binary string, process
> it's with array segment to get all informations then convert them to
> integer,
> to build the 32bit integer, i join all information value into binary
> string (yes, 32 bit of char ;p) join all of them then convert to
> integer.
>
> I know my way is sucks and too far away from COOL thing ;p,  i think
> there are cool way to do this!.
>
> Thanks before!
> --
> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
> the body of a message to majordomo@vger.kernel.org
> More majordomo info at  http://vger.kernel.org/majordomo-info.html
>



-- 
Uday
http://soundc.de/
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Re: Help on bit operation

[Ben Rosenberg]

On Wed, Aug 26, 2009 at 4:02 PM, Randi Botse<nightdecoder@gmail.com> wrote:
> Hi, I'm beginner C programmer, i have a problem, i want to store some
> information in a integer, a integer will be 32 bit on my machine, i
> want to have as follow:
>
> 6 bit (information 1)   MSB
> 4 bit (information 2)
> 8 bit (information 3)
> 5 bit (information 4)
> 9 bit (information 5)   LSB
>
> For example i set the informations as follow (in decimal):
>
> information 1 = 43        or 101011
> information 2 = 11        or 1011
> information 3 = 120      or 1111000
> information 3 = 30        or 11110
> information 4 = 418      or 110100010
>
> if i join all informations i should get a 32 bit integer valued
> 2935782212 or 01010111011111100011110110100010, then my problem is how
> to retrieve these informations on bit operation? i want to know what's
> the value of information-2 or information-3, etc directly.  And, is
> there any good way to join these informations to be an 32 bit integer?
>
> at this time i convert the 32bit integer into binary string, process
> it's with array segment to get all informations then convert them to
> integer,
> to build the 32bit integer, i join all information value into binary
> string (yes, 32 bit of char ;p) join all of them then convert to
> integer.
>
> I know my way is sucks and too far away from COOL thing ;p,  i think
> there are cool way to do this!.
>
> Thanks before!
> --
> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
> the body of a message to majordomo@vger.kernel.org
> More majordomo info at  http://vger.kernel.org/majordomo-info.html
>

A quick, dirty way of doing it without structs:

#include <stdio.h>

void store_in_int( unsigned int *target, unsigned int offset, size_t
field_sz, unsigned int value ) {
    unsigned int mask = ((1<<(field_sz))-1)<<offset;
    /* Clear field */
    *target |= mask;
    *target ^= mask;
    /* Load field */
    *target |= value<<offset;
}

unsigned int retrieve_from_int( unsigned int target, unsigned int
offset, size_t field_sz ) {
    unsigned int mask = ((1<<(field_sz))-1);
    return (target>>offset)&mask;
}

int main ( void ) {
    unsigned int info=0;
    store_in_int(&info, 0, 6, 43);
    store_in_int(&info, 6, 4, 11);
    store_in_int(&info,10, 8,120);
    store_in_int(&info,18, 5, 30);
    store_in_int(&info,23, 9,418);
    printf("info1: %u\n",retrieve_from_int( info, 0, 6 ));
    printf("info2: %u\n",retrieve_from_int( info, 6, 4 ));
    printf("info3: %u\n",retrieve_from_int( info,10, 8 ));
    printf("info4: %u\n",retrieve_from_int( info,18, 5 ));
    printf("info5: %u\n",retrieve_from_int( info,23, 9 ));
    return 0;
}
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Re: Help on bit operation

[Glynn Clements]

Randi Botse wrote:

> Hi, I'm beginner C programmer, i have a problem, i want to store some
> information in a integer, a integer will be 32 bit on my machine, i
> want to have as follow:
> 
> 6 bit (information 1)   MSB
> 4 bit (information 2)
> 8 bit (information 3)
> 5 bit (information 4)
> 9 bit (information 5)   LSB
> 
> For example i set the informations as follow (in decimal):
> 
> information 1 = 43        or 101011
> information 2 = 11        or 1011
> information 3 = 120      or 1111000
> information 3 = 30        or 11110
> information 4 = 418      or 110100010
> 
> if i join all informations i should get a 32 bit integer valued
> 2935782212 or 01010111011111100011110110100010, then my problem is how
> to retrieve these informations on bit operation? i want to know what's
> the value of information-2 or information-3, etc directly.  And, is
> there any good way to join these informations to be an 32 bit integer?
> 
> at this time i convert the 32bit integer into binary string, process
> it's with array segment to get all informations then convert them to
> integer,
> to build the 32bit integer, i join all information value into binary
> string (yes, 32 bit of char ;p) join all of them then convert to
> integer.
> 
> I know my way is sucks and too far away from COOL thing ;p,  i think
> there are cool way to do this!.

Two options:

1. Bit fields:

	struct information {
	#if __BYTE_ORDER == __LITTLE_ENDIAN
		unsigned int information_1 : 6;
		unsigned int information_2 : 4;
		unsigned int information_3 : 8;
		unsigned int information_4 : 5;
		unsigned int information_5 : 9;
	#else
		unsigned int information_5 : 9;
		unsigned int information_4 : 5;
		unsigned int information_3 : 8;
		unsigned int information_2 : 4;
		unsigned int information_1 : 6;
	#endif
	};

2. Shift and mask:

	value = value & ~0x3F  << 26 | (information_1 & 0x3F)  << 26;
	value = value & ~0x0F  << 22 | (information_2 & 0x0F)  << 22;
	value = value & ~0xFF  << 14 | (information_3 & 0xFF)  << 14;
	value = value & ~0x1F  <<  9 | (information_4 & 0x1F)  <<  9;
	value = value & ~0x1FF <<  0 | (information_5 & 0x1FF) <<  0;

	information_1 = value >> 26 & 0x3F;
	information_2 = value >> 22 & 0x0F;
	information_3 = value >> 14 & 0xFF;
	information_4 = value >>  9 & 0x1F;
	information_5 = value >>  0 & 0x1FF;

Regarding masks, the following should be memorised:

	Hex	Binary

	0	0000
	1	0001
	3	0011
	7	0111
	F	1111

	F	1111
	E	1110
	C	1100
	8	1000
	0	0000

-- 
Glynn Clements <glynn@gclements.plus.com>
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Re: Help on bit operation

[Randi Botse]

Thanks all for the clues!, that's so cool, i can now join and retrieve
informations on that 32bit integer with cool options that you told me.
 Thanks you very much!

Re: Help on bit operation

[Randi Botse]

Hi again.

I was able to solve my bit packing/unpacking problems with bit mask
and shift operations, but i have no idea how to do this using a
bit-field structure, can someone give me a example?

Thanks
             - Randi

Re: Help on bit operation

[Ben Rosenberg]

On Thu, Aug 27, 2009 at 9:52 PM, Randi Botse<nightdecoder@gmail.com> wrote:
> Hi again.
>
> I was able to solve my bit packing/unpacking problems with bit mask
> and shift operations, but i have no idea how to do this using a
> bit-field structure, can someone give me a example?
>
> Thanks
>             - Randi
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> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
> the body of a message to majordomo@vger.kernel.org
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>

Glynn's example of bitfields is cleaner and better than this, but I
put a basic implementation of bitfields using your parameters at the
end of this message. From my understanding, the way bitfields are
implemented in C is with bit shifts and masks, so I don't think there
is a performance difference between the two.

#include <stdio.h>

struct bitfield {
    unsigned int a:6;
    unsigned int b:4;
    unsigned int c:8;
    unsigned int d:5;
    unsigned int e:9;
};

int main(void) {
    struct bitfield foo;
    foo.a =  43;
    foo.b =  11;
    foo.c = 120;
    foo.d =  30;
    foo.e = 418;
    printf("%u %u %u %u %u\n",foo.a,foo.b,foo.c,foo.d,foo.e);
    return 0;
}


Ben
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Re: Help on bit operation

[Randi Botse]

Ben, after assign all bitfield struct member, how to pack them to be a
32bit integer value?

DEC      BIN
43        101011        -> foo.a (6 bit)
11        1011            -> foo.b (4 bit)
120      01111000    -> foo.c (8 bit)
30        11110          -> foo.d (5 bit)
418      110100010  -> foo.e (9 bit)

the bit pattern is : 10101110110111100011110110100010 or 2933800354 in
decimal, is this possible?

while with masking and shift i can pack them with:
((foo.a & 0x3f) << 26) | ((foo.b & 0xf) << 22) | ((foo,c & 0xff) <<
14) | ((foo.d & 0x1f) << 9) | (foo.e & 0x1ff)


On Sat, Aug 29, 2009 at 12:02 AM, Ben Rosenberg<scdlbx@gmail.com> wrote:

> Glynn's example of bitfields is cleaner and better than this, but I
> put a basic implementation of bitfields using your parameters at the
> end of this message. From my understanding, the way bitfields are
> implemented in C is with bit shifts and masks, so I don't think there
> is a performance difference between the two.
>
> #include <stdio.h>
>
> struct bitfield {
>    unsigned int a:6;
>    unsigned int b:4;
>    unsigned int c:8;
>    unsigned int d:5;
>    unsigned int e:9;
> };
>
> int main(void) {
>    struct bitfield foo;
>    foo.a =  43;
>    foo.b =  11;
>    foo.c = 120;
>    foo.d =  30;
>    foo.e = 418;
>    printf("%u %u %u %u %u\n",foo.a,foo.b,foo.c,foo.d,foo.e);
>    return 0;
> }
>
>
> Ben
>
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Re: Help on bit operation

[Ben Rosenberg]

On Fri, Aug 28, 2009 at 11:40 AM, Randi Botse<nightdecoder@gmail.com> wrote:
> Ben, after assign all bitfield struct member, how to pack them to be a
> 32bit integer value?
>
> DEC      BIN
> 43        101011        -> foo.a (6 bit)
> 11        1011            -> foo.b (4 bit)
> 120      01111000    -> foo.c (8 bit)
> 30        11110          -> foo.d (5 bit)
> 418      110100010  -> foo.e (9 bit)
>
> the bit pattern is : 10101110110111100011110110100010 or 2933800354 in
> decimal, is this possible?
>
> while with masking and shift i can pack them with:
> ((foo.a & 0x3f) << 26) | ((foo.b & 0xf) << 22) | ((foo,c & 0xff) <<
> 14) | ((foo.d & 0x1f) << 9) | (foo.e & 0x1ff)
>
>

The bitfield struct is already packed, you don't need to do anything
else. Just assign values to the different parts.

Ben
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Re: Help on bit operation

[Tim Walberg]

[-- Attachment #1: Type: text/plain, Size: 1747 bytes --]

If you absolutely need to get at the 32-bit integer value, you can use

union mytype {
{ struct bits {
    unsigned int a:6;
    unsigned int b:4;
    unsigned int c:8;
    unsigned int d:5;
    unsigned int e:9;
  } b;
  unsigned int v;
} variable;

then access the bit-fields like

  unsigned int x = variable.b.e;
  variable.b.a = 3;
  x = variable.v;


  


On 08/28/2009 12:58 -0700, Ben Rosenberg wrote:
>>	On Fri, Aug 28, 2009 at 11:40 AM, Randi Botse<nightdecoder@gmail.com> wrote:
>>	> Ben, after assign all bitfield struct member, how to pack them to be a
>>	> 32bit integer value?
>>	>
>>	> DEC      BIN
>>	> 43        101011        -> foo.a (6 bit)
>>	> 11        1011            -> foo.b (4 bit)
>>	> 120      01111000    -> foo.c (8 bit)
>>	> 30        11110          -> foo.d (5 bit)
>>	> 418      110100010  -> foo.e (9 bit)
>>	>
>>	> the bit pattern is : 10101110110111100011110110100010 or 2933800354 in
>>	> decimal, is this possible?
>>	>
>>	> while with masking and shift i can pack them with:
>>	> ((foo.a & 0x3f) << 26) | ((foo.b & 0xf) << 22) | ((foo,c & 0xff) <<
>>	> 14) | ((foo.d & 0x1f) << 9) | (foo.e & 0x1ff)
>>	>
>>	>
>>	
>>	The bitfield struct is already packed, you don't need to do anything
>>	else. Just assign values to the different parts.
>>	
>>	Ben
>>	--
>>	To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
>>	the body of a message to majordomo@vger.kernel.org
>>	More majordomo info at  http://vger.kernel.org/majordomo-info.html
End of included message



-- 
+----------------------+
| Tim Walberg          |
| 830 Carriage Dr.     |
| Algonquin, IL 60102  |
| twalberg@comcast.net |
+----------------------+

[-- Attachment #2: Type: application/pgp-signature, Size: 197 bytes --]

Re: Help on bit operation

[Glynn Clements]

Randi Botse wrote:

> Ben, after assign all bitfield struct member, how to pack them to be a
> 32bit integer value?

The structure should be 32 bits wide. If you need the value as a
uint32_t, either use a union:

	union bits_or_int {
		struct bitfield fields;
		uint32_t value;
	};

or a cast, e.g.:

	struct bitfield foo;
	...
	uint32_t value = *(uint32_t *)&foo;

-- 
Glynn Clements <glynn@gclements.plus.com>

Re: Help on bit operation

[Randi Botse]

everything is clear now!, thanks all for your guides and patience.

    - Randi


On Sat, Aug 29, 2009 at 3:19 AM, Glynn Clements<glynn@gclements.plus.com> wrote:
>
> Randi Botse wrote:
>
>> Ben, after assign all bitfield struct member, how to pack them to be a
>> 32bit integer value?
>
> The structure should be 32 bits wide. If you need the value as a
> uint32_t, either use a union:
>
>        union bits_or_int {
>                struct bitfield fields;
>                uint32_t value;
>        };
>
> or a cast, e.g.:
>
>        struct bitfield foo;
>        ...
>        uint32_t value = *(uint32_t *)&foo;
>
> --
> Glynn Clements <glynn@gclements.plus.com>
>
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