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cfunctionvariablestypedef
darksky
darksky

Reputation: 21019

Typedef Function Syntax

I have the following typedef function prototype:

typedef void (*sa_sigaction_t)(int, siginfo_t *, void *);

I have no idea how to use it. I understand that (int, siginfo_t *, void*) is typedef-ed into sa_sigaction_t. But how would I use it? These are all return types with no variable names specified.

So I assume I want to create a variable: sa_sigaction_t handler. How would I use it? The syntax is quite confusing.

Thanks,

Upvotes: 1

Views: 890

Answers (3)

Mike Kwan
Mike Kwan

Reputation: 24447

The typedef in this case is sort of a short hand for a function which returns void and takes 3 arguments (int, siginfo_t * and void *). This is mainly useful for if you want to pass a function around as a callback for example.

void func1(sa_sigaction_t handler)
{
  handler(...);
}

func1 invokes any function of type sa_sigaction_t. You can call it like so:

void func2(int a, siginfo_t * b, void * c)
{
  ...
}

int main(void)
{
  func1(func2);

  /*
   * Equivalent to:
   * sa_sigaction_t handler = func2;
   * func1(func2);
   */
  return 0;
}

On the other hand, if you did not have the typedef, your code would be more verbose:

void func1(void(*handler)(int, siginfo_t *, void*))
{
  handler(...);
}

void func2(int a, siginfo_t * b, void * c)
{
  ...
}

int main(void)
{
  func1(func2);

  /*
   * Equivalent to:
   * void(*handler)(int, siginfo_t *, void*) = func2;
   * func1(func2);
   */
  return 0;
}

Something to note is that sometimes you will see func2 and other times &func2, but these are both the same thing.

Upvotes: 1

Luchian Grigore
Luchian Grigore

Reputation: 258568

I understand that (int, siginfo_t , void) is typedef-ed into sa_sigaction_t.

Actually no. sa_sigaction_t is a pointer to a function that returns void and takes (int, siginfo_t *, void *) as parameters.

So if you have:

void foo(int, siginfo_t*, void*)
{
}

you can do:

sa_sigaction_t fooPtr = &foo;

and then call it like this:

fooPtr(0,NULL,NULL);

Upvotes: 1

James M
James M

Reputation: 16718

If you have a function pointer declared:

sa_sigaction_t handler;

You can call it like:

handler( ... );

The clue is in the brackets, really. void (*sa_sigaction_t) is different from void *sa_sigaction_t .

Upvotes: 0

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