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<<|>>Bit packing

Posted on 1998-09-18
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Last Modified: 2010-04-15
Hi,


I am trying to pack 4 6 bits data into 3 bytes. How do I check whether the packing that I have done is correct? I don't want to write the codes for unpacking the data before checking that the codes for packing is correct. Following is my code segment :

int packbit (unsigned char *a, int n, unsigned char *b)
{
  int i;
  int loop=n/4;
 
for(i=0; i<loop;a+=4,b+=3)
   {
      b[0]=a[0]+(unsigned char)(a[1]<<6);
      b[1]=(a[1]>>2)+(unsigned char)(a[2]<<4);
      b[2]=(a[2]>>4)+(unsigned char)(a[3]<<2);
   }
}

Pls help with some code examples. Thanks.

Rgds,
PY
0
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Question by:peiyoke
2 Comments
 
LVL 4

Accepted Solution

by:
emmons earned 10 total points
ID: 1252874
First, you have a iterator in the for loop that is never incremented, that causes something of a problem when the code is run.
Second, you seem to be trying to make the pack routine more general than the description in your message, and yet there is an assumption in the code that it is always packing 4 bytes into 3.

To do the conversion, try
      b[0] = ((a[0] << 2) & 0xFC) + ((a[1] >> 4) & 0x03);
      b[1] = ((a[1] << 4) & 0xF0) + ((a[2] >> 4) & 0X0F);
      b[2] = ((a[2] << 6) & 0xF0) + ((a[3] & 0x3F));

0
 

Author Comment

by:peiyoke
ID: 1252875
What I am trying to do is to compress data in a file and packing algorithm to be used is not restricted. After running the codes for packing, can I determine from output directly whether the codes is correct? I tried to examine the bit pattern of the output but don't see any correlation between the input and the output data.

Even though I store my data as unsigned char, I came across a book which states that integral promotions are performed when shift expressions are evaluated. Does this means that even though my data are packed from 4 bytes to 3 bytes for the array elements, but the overall storage required to store the data has been increased by 4?

Can I or should I do some casting such as
(unsigned char)(a[0]>>2)?


Pls advise. Thanks.

0

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