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@xxxxxxxxxxxxxxxxxx>
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