UNIX for Advanced & Expert Users

I have a program............

Code:

#include<stdio.h>
#include<unistd.h>
main()
{
if(fork == 0)
{
printf("Hi every body:p!!!!!!!!!!");
}
}

This program works with out any error. here fork is not a system call. It just act as a variable.But how it works without declaring it? What data type it is?

Please help me to get a solution. I have tried all . But i cant come out with an answer.

you remove the #include<unistd.h> header file then see what error it is saying.

see the following example.
in aa.c file content is

Code:

int ss=2;

then I include the file in the following program.

Code:

#include<stdio.h>
#include "aa.c"
main()
{
        printf("%d",ss);
}

it will work. but I didn't declare the ss in this file.

how it's work? did you understand this example with your code.

the fork variable is declared in #include<unistd.h> file. so it's not showing the error.

if you are using vim. editor then place the cursor on the "#include<unistd.h> " and hit gf then the "unistd.h" file will open. then search the variable fork.

But i dint find any declared variable like fork in unistd.h. In your example i can tell ss is an integer type by seeing the include header file..but here can you tell wat datatype fork is? And in a declared variable i can do ++ or -- operation. but here its not working.

No fork in unistd ?

Code:

$ grep -w fork /usr/include/unistd.h
extern pid_t fork(void);
extern pid_t fork();

her fork is not a function or system call

so what ?

fork in this case is not a variable. It's a pointer of type void to the entry address of the fork() function in the C library. Basically, you could "rename" the fork call by doing

Code:

__pid_t (*foo)() = fork;

and later calling foo instead of fork.

Also, if you'd have compiled with warnings enabled, or read them, you'd have seen something like this:

Code:

warning: the address of ‘fork’ will never be NULL