Hi,
I want a generate a node that consists of three integers but I want to store these three integers in bits and not bytes
e.g if the node has the values 4,3,2
in integers it will be stored as 00000100-00000011-00000010
but if I store them in bits then they will be 01001110 occupy 1 byte instead of three
taking into consideration that the values are undetermined in the compilation time(I can’t use byte: x 4)
can u tell me how can I do such a thing and how can I dump them as 4,3& 2 again
thx for time

All 9 Replies

``````#include <stdio.h>
#include <limits.h>

struct type /* note: bit order is implementation-dependent */
{
unsigned int a : 2;
unsigned int b : 2;
unsigned int c : 4;
};

void showbits(unsigned char byte)
{
unsigned char bit;
for ( bit = 1 << (CHAR_BIT - 1); bit; bit >>= 1 )
{
putchar(byte & bit ? '1' : '0');
}
putchar('\n');
}

int main()
{
struct type value = {2,3,4};
printf("sizeof value = %d\n", (int)(sizeof value));
printf("value.a = %u\n", value.a);
printf("value.b = %u\n", value.b);
printf("value.c = %u\n", value.c);
showbits(*(unsigned char*)&value);
return 0;
}

/* my output
sizeof value = 1
value.a = 2
value.b = 3
value.c = 4
01001110
*/``````

Greetings,

Binary operations can be somewhat confusing at times, though quite simple to understand. It is customary to refer to a digit in binary (either 1 or 0) as a bit, likewise a byte is a set of eight bits. So, for example, 10100110 is a byte. A "nibble" is half of a byte, as it is a binary number with four bits.

A word is a sixteen-bit binary number. It is equivalent to two bytes, or four nibbles. A double word is a 32-bit binary number, so a dword is equivalent to two words, or four bytes, or eight nibbles. Most commonly, the number of bits (or bytes) in a number-type is referred to as that type's width.

Bit-type Examples:

``````Bit		1 bin
Nibble		0110 bin
Byte		00100101 bin
Word		1110100001010110 bin
D Word	 	10100110000111101110111011010101 bin``````

More information about Binary and its essentials can be found here.

- Stack Overflow

This is perhaps a more portable way than my previous post.

``````#include <stdio.h>
#include <limits.h>

void showbits(unsigned char byte)
{
unsigned char bit;
for ( bit = 1 << (CHAR_BIT - 1); bit; bit >>= 1 )
{
putchar(byte & bit ? '1' : '0');
}
putchar('\n');
}

unsigned char pack(unsigned char a, unsigned char b, unsigned char c)
{
return ((a & 0xF) << 4) | ((b & 0x3) << 2) | (c & 0x3);
}

void unpack(unsigned char value, unsigned char *a, unsigned char *b, unsigned char *c)
{
*a = (value >> 4) & 0xF;
*b = (value >> 2) & 0x3;
*c =  value	   & 0x3;
}

int main()
{
unsigned char a, b, c, value = pack(4,3,2);
printf("sizeof value = %d\n", (int)(sizeof value));
showbits(value);
unpack(value, &a, &b, &c);
printf("a = %d, b = %d, c = %d\n", a, b, c);
return 0;
}

/* my output
sizeof value = 1
01001110
a = 4, b = 3, c = 2
*/``````

Binary operations can be somewhat confusing at times, though quite simple to understand. It is customary to refer to a digit in binary (either 1 or 0) as a bit, likewise a byte is a set of eight bits. So, for example, 10100110 is a byte. A "nibble" is half of a byte, as it is a binary number with four bits.

A word is a sixteen-bit binary number. It is equivalent to two bytes, or four nibbles. A double word is a 32-bit binary number, so a dword is equivalent to two words, or four bytes, or eight nibbles. Most commonly, the number of bits (or bytes) in a number-type is referred to as that type's width.

In C and C++ these identities do not necessarily hold. There can be a 16-bit byte and a 1-byte int, and other such configurations.

This is perhaps a more portable way than my previous post.

``````#include <stdio.h>
#include <limits.h>

void showbits(unsigned char byte)
{
unsigned char bit;
for ( bit = 1 << (CHAR_BIT - 1); bit; bit >>= 1 )
{
putchar(byte & bit ? '1' : '0');
}
putchar('\n');
}

unsigned char pack(unsigned char a, unsigned char b, unsigned char c)
{
return ((a & 0xF) << 4) | ((b & 0x3) << 2) | (c & 0x3);
}

void unpack(unsigned char value, unsigned char *a, unsigned char *b, unsigned char *c)
{
*a = (value >> 4) & 0xF;
*b = (value >> 2) & 0x3;
*c =  value	   & 0x3;
}

int main()
{
unsigned char a, b, c, value = pack(4,3,2);
printf("sizeof value = %d\n", (int)(sizeof value));
showbits(value);
unpack(value, &a, &b, &c);
printf("a = %d, b = %d, c = %d\n", a, b, c);
return 0;
}

/* my output
sizeof value = 1
01001110
a = 4, b = 3, c = 2
*/``````

thx for ur help, but all this is under the assumption that the first char must be packed in 4 bits, the second in 2bits,.. but what if the sizes increase ,what about if i want to write 200,400,3 in bits how can i do so

thx for ur help, but all this is under the assumption that the first char must be packed in 4 bits, the second in 2bits,.. but what if the sizes increase ,what about if i want to write 200,400,3 in bits how can i do so

Well, if you want to store a nine-bit value (400) in an eight-bit object (the common unsigned char), you are going to have issues now, aren't you?

Well, if you want to store a nine-bit value (400) in an eight-bit object (the common unsigned char), you are going to have issues now, aren't you?

yeah it looks so , but how can i solve such a problem, i'm sure there is a solution
actually i can check for the long of the number in binary and make shifting the number dynamically but in such away the problem will be in retreiving the numbers back .
what do u think ;)

Do your values have a known fixed range? Then the solutions I presented can be modified.

Or are you trying to compress data? Then search for compression algorithms.

Or can each value be of a different size? ASN.1 and BER is a way to handle that one.

What is the "big picture"?

Do your values have a known fixed range? Then the solutions I presented can be modified.

Or are you trying to compress data? Then search for compression algorithms.

Or can each value be of a different size? ASN.1 and BER is a way to handle that one.

What is the "big picture"?

actually i'm trying to compress data but i can't find an answer in compression algorithms
values are of no fixed size ,they can be anything but one of them muct always take 7bits(unsigned char)the other two must be packed in bits
so what do u think??????? :sad:

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