Source in C for longitudinal redundancy check?

Answer 1

#include<stdio.h> #include<conio.h>

int binary(int);

void parity(int [],int []);

int arr[8],arr1[8],parityarr[8];

char chr;

int go,temp,temp1,i;

void main()

{ char chr1;

clrscr();

go=0;

Sender:

if(go==0)

printf("Enter Data : \n");

printf("\nEnter a character : ");

scanf("%c %c",&chr,&chr1);

temp=chr;

binary(temp);

printf("\n\nAscii value is : %d\n",temp);

printf("\nBinary Form : ");

for(i=7;i>=0;i--)

{

arr1[i]=arr[i];

printf("%d ",arr[i]);

}

printf("\n");

temp1=chr1;

binary(temp1);

printf("\n\nAscii value is : %d\n",temp1);

printf("\nBinary Form : ");

for(i=7;i>=0;i--)

{

printf("%d ",arr[i]);

}

parity(arr,arr1);

for(i=7;i>=0;i--)

{

arr[i]=0;

arr1[i]=0;

}

getch();

}

void parity(int arr[],int arr1[])

{

printf("Receiver Side :\n");

printf("\n\nLRC :\n");

for(i=7;i>=0;i--)

{

printf("%d ",arr[i]);

}

printf("\n");

for(i=7;i>=0;i--)

{

printf("%d ",arr1[i]);

}

printf("\n---------------------\n");

for(i=0;i<8;i++)

{

if(arr[i]==0 && arr1[i]==0 arr[i]==1 && arr1[i]==1)

{

parityarr[i]=0;

}

else if(arr[i]==1 && arr1[i]==0 arr[i]==0 && arr1[i]==1)

{

parityarr[i]=1;

}

}

for(i=7;i>=0;i--)

{

printf("%d ",parityarr[i]);

}

}

int binary(int x)

{

int rem;

int ctr=0,i=1;

do

{

rem=x%2;

arr[i]=rem;

if(rem==1)

{

ctr++;

}

x=x/2;

i++;

}

while(x!=0);

if(ctr%2==0)

{

arr[0]=0;

}

else

{

arr[0]=1;

}

return(0);

}

Answer 2

Here is a simple example of Longitudinal Redundancy Check (LRC) implementation in C:

#include <stdio.h>

unsigned char calculate_lrc(unsigned char *data, int length) {
    unsigned char lrc = 0x00;
    for (int i = 0; i < length; i++) {
        lrc ^= data[i];
    }
    return lrc;
}

int main() {
    unsigned char data[] = {0x01, 0x02, 0x03, 0x04};
    int length = sizeof(data) / sizeof(data[0]);
    unsigned char lrc = calculate_lrc(data, length);
    printf("LRC: 0x%02X\n", lrc);
    return 0;
}