[SOLVED] I need help understanding this declaration in C...

trist007 · 07-24-2011, 01:10 PM

Code:

           struct sigaction {
               void     (*sa_handler)(int);
               void     (*sa_sigaction)(int, siginfo_t *, void *);
               sigset_t   sa_mask;
               int        sa_flags;
               void     (*sa_restorer)(void);
           };

On the 2nd line

Code:

void (*sa_handler)(int);

Is the line declaring a void pointer with size of int?

Nylex · 07-24-2011, 01:49 PM

No. It's a pointer to a function that returns void and takes an int as an argument. For example, if you had a function

void foo(int x) { // Do stuff },

you could write

struct sigaction sa;
sa.sa_handler = &foo;,

where & is the address of operator.

trist007 · 07-24-2011, 02:45 PM

It's a pointer to which function?

Nylex · 07-24-2011, 02:46 PM

Any function that returns void and takes an int as an argument. The pointer hasn't been given an address yet, so it's currently null.

Edit: In the example I gave above, sa_handler would be a pointer to the function foo().

trist007 · 07-24-2011, 03:19 PM

Interesting, never seen it done this way, but ok. Thanks for explaining Nylex.

Nylex · 07-24-2011, 03:22 PM

You haven't seen pointers to functions before? I can't say I really know what they're for, as I've never used them myself

. Hopefully someone else will provide a better explanation!

trist007 · 07-24-2011, 03:31 PM

So in a case like this.

Code:

        // ignore SIGPIPE
        struct sigatction act;
        sigaction(SIGPIPE, (struct sigaction *)NULL, &act);  // grabs old action
        if(act.sa_handler == SIG_DFL)           // if default
        {
                act.sa_handler == SIG_IGN;   // sets handler to ignore
                sigaction(SIGPIPE, &act, (struct sigaction *)NULL); // puts in new action
        }

Code:

act.sa_handler == SIG_IGN;    // which sets it to ignore

SIG_IGN is also an int.
Trying to tie in

Code:

void (*sa_handler)(int)

in this situation. How does this apply?

markush · 07-24-2011, 04:26 PM

Hello,

you may take a look at the "function-pointer tutorial" http://www.newty.de/fpt/index.html
it has some useful examples.

Btw, I've always found this pointers to be an interesting concept, but I'm not really a C-programmer and so I never have used it.

Markus

trist007 · 07-24-2011, 04:44 PM

Perfect, I will read up, and then come back and hopefully be able to explain what's going on above as so as to help others that come across my post. Thanks again.

Nominal Animal · 07-24-2011, 04:47 PM

Code:

void (*something)(int);

defines something as a pointer to a function, which returns nothing (void), but takes one integer as a parameter (int).

The way you use function pointers is simple; you just assign the pointer, then use it as if it were a function. Consider this example:

Code:

#include <stdio.h>

int (*operation)(int, int);

int addition(int a, int b) { return a + b; }
int substraction(int a, int b) { return a - b; }

int main(void)
{
    printf("operation = addition;\n");
    operation = addition;

    printf("/* operation(1, 2) == %d */\n", operation(1, 2));

    printf("\n");

    printf("operation = substraction;\n");
    operation = substraction;

    printf("/* operation(1, 2) == %d */\n", operation(1, 2));
    return 0;
}

You can, but do not need to use the & operator in front of a function name, as long as you omit the parentheses (). The compiler knows you mean the address of the function in any case. I prefer to not use it, so the compiler will warn me if the target is not a function (or a function pointer). Also, you cannot use the & when the target is a function pointer (because then you'd take the address of the pointer, not the target of the pointer). So I generally recommend to not use the & operator when taking the address of a function.

When compiled and run, the program will output

Code:

operation = addition;
/* operation(1, 2) == 3 */

operation = substraction;
/* operation(1, 2) == -1 */

When defined in a structure, function pointers behave exactly the same. For example,

Code:

           struct sigaction {
               void     (*sa_handler)(int);
               void     (*sa_sigaction)(int, siginfo_t *, void *);
               sigset_t   sa_mask;
               int        sa_flags;
               void     (*sa_restorer)(void);
           };

defines three function pointers as fields in the structure:

sa_handler is a pointer to a function which takes one int as a parameter, and returns nothing (void).
sa_sigaction is a pointer to a function which takes three parameters: an int, a pointer to a siginfo_t (which happens to be a structure), and a void pointer.
sa_restorer is a pointer to a function which takes no parameters (void), and returns nothing (void).

The sigaction() function call understands two constants for sa_handler: SIG_IGN and SIG_DFL . They are both pointers to a function that takes one int as a parameter, and returns nothing.

Note that SIG_DFL and SIG_IGN are just pointers, not real functions. For a number of reasons, they are defined to point to specific addresses (in Linux, to addresses 0 and 1, respectively); just like NULL points to address 0. So, while you could call SIG_IGN(SIGUSR1) or SIG_DFL(SIGHUP), doing that will crash your program since SIG_IGN and SIG_DFL do not point to any real functions.

Hope this clarifies the issue for you.

sundialsvcs · 07-24-2011, 07:43 PM

There are many peculiar things that you will find in "C" code. It is a language designed to be expressive of very low-level constructs when necessary.

trist007 · 07-25-2011, 08:46 AM

Thanks a lot that cleared it up.