Programming

hi folks!
how to deamonize my program,
and also here is a code that my collegue asked me which i could not explain

int main()
{
char **p = &p;
printf("%s\n",*p + 40);
return 0;
}

after compling this gives a warning,
and the output is something out of the sun,can u tell me why it is so.

To answer your questions:
1. How to daemonize your program: this is a very nice explaination of how do that.

2. Regarding the warning: You are getting that cause the program is trying to assign an address (probably an int, definitely numerical) value to a char type of variable. You can write the code as

Code:

char **p=(char**)&p;

and prevent the warning from being thrown.

3. Regarding the output: You are printing an address, which is some random number, as a string. What output are you expecting? Try printing that as an integer.

That was a very good link blowtorch.

I think the warning is not because an int type is assigned to a char type ( infact, p when declared as char **p; is definetly not a character variable). The warning comes because of the fact that we are assigning address of a pointer variable to a pointer-to-pointer type variable, and by doing a typecast

Code:

char **p = (char **)&p;

u r typecasting a pointer ( &p) to a pointer-to-pointer ...

Infact, this will also work ...

Code:

char **p = *(&p);

Try doing this on paper with a pen ...

and yes blowtorch, that was a very good link... I too found it useful

The code shown in that link will allow the deamon to remain a session group leader. If a daemon who is a session group leader opens a tty device without specifying the O_NOCTTY flag, the daemon will un-daemonize itself. That is why I prefer: 1.7 How do I get my program to act like a daemon?