2 replies to this topic

[ChilledOut]

I just learned about this today. There are a couple of things I don't completely understand, but to avoid overwhelming myself and other people on this forum I'm going to take it one step at a time.

The first thing I'm having trouble with is making a structure into a function. I know the function call should look like this:

database fn();

Where database is the type of struct and fn is the function name. What I don't know is what return type to make the function. I have read that it should be the same return type as the structure itself but that hasn't worked for me yet.

After this is solved I will move on to the next, less serious thing.

[WingedPanther]

You've just declared that the return type of fn() is database.

[Alexander]

If I get what you are doing, you could always return i.e. type database* to use in your calling code.

i.e.

database* fn() {
  struct database* mydb = malloc(sizeof(struct database));
  if(mydb != NULL) {
    mydb->id = 12;
  }
  return mydb;
}

..
struct database *caller = NULL;
caller = fn();

You could as well manipulate caller through a pointer, i.e. defining

void fn(struct database* mydb) {
  if(mydb != NULL) {
    mydb->id = 21;
  }
}
..
struct database *caller = NULL;
fn(&caller);

It should be better to do this, if you could provide a bit more info feel free to ask as always.

Alexander.