is it possible to pass arguments to the clone fn function?

hello!
I've a problem with the clone.. Its definition is as follows:

Code:

 int clone(int (*fn)(void *), void *child_stack, int flags, void *arg, .../* pid_t *pid, struct user_desc *tls, pid_t *ctid */ );

if I define a simple

Code:

 int function(void *unused)

everything's ok: i can call for example

Code:

clone ((int (*)(void *)function,&stack[1024],....

or also just

Code:

clone (function,&stack....

the problem is if i want to cast the void* to, e.g. an int*
i tried to do things like

Code:

int function (int *nPtr){...}
int a;
clone((int(*)(int *)function(&a),....
clone (function ((int *)&a), ...

but nothing works... where is the mistake? thank for every answer!

You have a 4th argument to clone called "void *arg". This is passed unchanged along to the function you specify. Which by the way should be specified like this:

Code:

clone (function,&stack, ...

You should NEVER use casts [unless you REALLY ABSOLUTELY know what you are doing] on function pointer arguments.

--
Mats

You don't do that. function must keep its prototype. If it wants to cast, you cast inside of function.

The first argument expects a signature of "int (*fn)(void *)" - that's exactly what you should give it. Any parameter interpretations should be done within the function:

Code:

int foo(void *p)
{
    int param = *((int*)p); // address of int passed
    // or 
    int param = (int)p; // int passed, cast as void*
}

// edit - of course I type this up and there's two replies already

gg

thank you guys!