mplementing hamming code in c lang
 

hey will u pls try to help me implementing hamming code in c:

Calculating the Hamming Code
The key to the Hamming Code is the use of extra parity bits to allow the identification of a single error. Create the code word as follows:

Mark all bit positions that are powers of two as parity bits. (positions 1, 2, 4, 8, 16, 32, 64, etc.)
All other bit positions are for the data to be encoded. (positions 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, etc.)
Each parity bit calculates the parity for some of the bits in the code word. The position of the parity bit determines the sequence of bits that it alternately checks and skips.
Position 1: check 1 bit, skip 1 bit, check 1 bit, skip 1 bit, etc. (1,3,5,7,9,11,13,15,...)
Position 2: check 2 bits, skip 2 bits, check 2 bits, skip 2 bits, etc. (2,3,6,7,10,11,14,15,...)
Position 4: check 4 bits, skip 4 bits, check 4 bits, skip 4 bits, etc. (4,5,6,7,12,13,14,15,20,21,22,23,...)
Position 8: check 8 bits, skip 8 bits, check 8 bits, skip 8 bits, etc. (8-15,24-31,40-47,...)
Position 16: check 16 bits, skip 16 bits, check 16 bits, skip 16 bits, etc. (16-31,48-63,80-95,...)
Position 32: check 32 bits, skip 32 bits, check 32 bits, skip 32 bits, etc. (32-63,96-127,160-191,...)
etc.
Set a parity bit to 1 if the total number of ones in the positions it checks is odd. Set a parity bit to 0 if the total number of ones in the positions it checks is even.
Here is an example:

A byte of data: 10011010(ip-user)
c programming for follwing steps:
Create the data word, leaving spaces for the parity bits: _ _ 1 _ 0 0 1 _ 1 0 1 0
Calculate the parity for each parity bit (a ? represents the bit position being set):

Position 1 checks bits 1,3,5,7,9,11:
? _ 1 _ 0 0 1 _ 1 0 1 0. Even parity so set position 1 to a 0: 0 _ 1 _ 0 0 1 _ 1 0 1 0
Position 2 checks bits 2,3,6,7,10,11:
0 ? 1 _ 0 0 1 _ 1 0 1 0. Odd parity so set position 2 to a 1: 0 1 1 _ 0 0 1 _ 1 0 1 0
Position 4 checks bits 4,5,6,7,12:
0 1 1 ? 0 0 1 _ 1 0 1 0. Odd parity so set position 4 to a 1: 0 1 1 1 0 0 1 _ 1 0 1 0
Position 8 checks bits 8,9,10,11,12:
0 1 1 1 0 0 1 ? 1 0 1 0. Even parity so set position 8 to a 0: 0 1 1 1 0 0 1 0 1 0 1 0
Code word: 011100101010. (output)

So you copied out your homework question word-for-word into a forum post. What do you expect us to do with that?

Narrow down what you need help with. How do you do that? Well, first you do the homework. If you end up with a solution that doesn't quite work, then you'll be able to post here about the part that's not working.

Giving you simple implementation. But seriously , Try it yourself. Google is your friend and wiki is your counseller.
//hamminggen.c
#include<stdio.h>
void main()
{
int d[4],c[7],i;
printf("\n Enter the 4 bit date seperated by blanks\n");
for(i=0;i<4;i++)
scanf("%d",&d[i]);
for(i=0;i<4;i++)
c[i]=d[i];
c[4]=(d[0]+d[2]+d[3])%2;
c[5]=(d[0]+d[1]+d[3])%2;
c[6]=(d[1]+d[2]+d[3])%2;
printf("\n The data bit appended with correction bit is \n");
for(i=0;i<7;i++)
printf("%d",c[i]);
}





// hammingdetection.c
#include<stdio.h>
void main()
{
int r[7],s[3],index,i,j,sum;
int h[3][7]={{1,0,1,1,1,0,0},{1,1,0,1,0,1,0},{0,1,1,1,0,0,1}};
printf("\n Enter the 7 bit information: \n");
for(i=0;i<7;i++)
scanf("%d",&r[i]);
for(j=0;j<3;j++)
{
sum=0;
for(i=0;i<7;i++)
sum+=h[i][j]*r[i];
sum=sum%2;
s[j]=sum;
}
if(s[0]==0 && s[1]==0 && s[2]==0)
{
printf("\n Error free Implementation.\n");
printf("\n The data bits are\n");
for(i=0;i<4;i++)
printf("%d",r[i]);
exit(0);
}
for(j=0;j<7;j++)
{
if(s[0]==h[0][j] && s[1]==h[1][j] && s[2]==h[2][j])
{
index=j;
break;
}
}
printf("\n The error is in bit no %d\n",index+1);
if(r[index]==0)
r[index]=1;
else
r[index]=0;
printf("\n The correct information is :");
for(i=0;i<7;i++)
printf("%d",r[i]);
printf("\n The data bits are : ");
for(i=0;i<4;i++)
printf("%d",r[i]);
}