Function Pointers within structures in C

Question

typedef struct Item{

  int i;
  int j;    
  void (*fooprint)(item*);

}item;

void fooprint(item *it){
  printf("%d\n",it.i);
}


int main(){

  item myitem;
  myitem.i=10;
  myitem.j=20;
  myitem.fooprint = fooprint;

  myitem.fooprint(&myitem);
  return 0;
}

This code gives a error at void (footprint)(item). "The error is expected ‘)’ before ‘*’ token ". Am I missing something ? When I do the same without using pointer to the structure is works. Example : void (*footprint)(item)

Johannes Weiss · Accepted Answer · 2013-11-11 00:33:14Z

7

The type item is not known yet when you use it. You can solve that with a forward declaration.

typedef struct Item item;
struct Item {
    int i;
    int j;
    void (*fooprint)(item*);
};

Another possibility is not to use the typedef to define members:

typedef struct Item {
    int i;
    int j;
    void (*fooprint)(struct Item *);
} item;

answered Nov 11, 2013 at 0:33

Johannes Weiss

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3 Comments

LeatherFace Over a year ago

That's what I was thinking, and it through me off when he said it works when he leaves the asterisk off.

Pavan Over a year ago

Now, the warning I have is :"useless storage class specifier in empty declaration"

Johannes Weiss Over a year ago

Yeah, it doesn't recognise item as a type (because it's not defined yet). Your compiler thinks here that it is a storage declaration (usually auto, register, ...) for an empty type. But that's just an implementation detail of your compiler. My compiler (clang) outputs error: a parameter list without types is only allowed in a function definition

alecbz · Accepted Answer · 2013-11-11 00:37:09Z

5

I'm not sure why you're getting the particular error you are -- the error I got was "error: unknown type name ‘item’". This is because the typedef has not "happened" yet, and C doesn't know what the type item refers to. Use struct Item in place of item there.

(Also, it.i in the printf should be it->i).

answered Nov 11, 2013 at 0:37

alecbz

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