I wanted to ask that how can we use Lables for calling a predefined kernel function from another function.

For .eg..

ssize_t scull_write()

{

write: xyz;

}

ssize scull_read()

{

goto write;

}

this calling of the write function is not working.

Could you please tell me that how can i get back to the write func from read func without calling it externally.

Thank you

#ifndef _IOC_WRITE
# define _IOC_WRITE     1U
#endif

#ifndef _IOC_READ
# define _IOC_READ      2U
#endif

please explain these lines of code>>!!

#ifndef _IOC_WRITE
# define _IOC_WRITE 1U
#endif

#ifndef _IOC_READ
# define _IOC_READ 2U
#endif

please explain these MACROS and their Scalar values used for determining the direction

Fedora installation

Kernel compilation

Gcc installation

Configure Raid

all servers

Created Usb media

Dual Boot

Debiyan installation

#include
int main(void)
{

struct node
{
int info;
struct node *d;
};
struct node d1,*n,*t;

{ int i;
d1.d=(struct node*) malloc(sizeof(struct node));
printf(“\nmemory allocation for pointer p of d1\n”);

printf(“\nd1.d=%d\n”,d1.d);
n=(struct node*)malloc(sizeof(struct node));
printf(“\ngiving base address to pointer n\n”);

printf(“\n*n=%d\n”,n);
d1.d=n;
printf(“\n2nd node created.pointer d is having base address of pointer n\n”);

printf(“\nd1.d=%d”,d1.d);

t=n;
printf(“\nlinking 2nd 3rd node t=n\n”);

printf(“\n*t=%d\n”,*t);
for(i=0;id=n;
printf(“\nlinking node %d %d =%d\n”,i+3,i+4,t->d);
t=(struct node*)malloc(sizeof(struct node));
printf(“\n getting anothrer base adderess =%d\n”,*t);

n->d=t;
printf(“\nlinking node %d %d =%d\n”,i+4,i+5,n->d);

} return 0;

————————————————————————————————————————————————

output for linked list……………………….

memory allocation for pointer p of d1

d1.d=149512200

giving base address to pointer n

*n=149512216

2nd node created.pointer d is having base address of pointer n

d1.d=149512216
linking 2nd 3rd node t=n

*t=0

the loop has started from here

n has got the base adderess=0

linking node 3 4 =149512232

getting anothrer base adderess =0

linking node 4 5 =149512248

the loop has started from here

n has got the base adderess=0

linking node 4 5 =149512264

getting anothrer base adderess =0

linking node 5 6 =149512280
………………………………………………………………………………….

sir as i have taken 2 pointers..n equate them in 2/3rd step so that pointers should have same addresses bt after equating them gettin *t=0…..

kindly tell me the areas at which im getting short…..
thank you