void pointer to function

Question

I need an explanaton. I searched for an answer but couldn't find one.

The problem is as follows: I'm working on exercises from the book: "Programming for Engineers" A.R. Bradley Chapter 7 Here's part of the code:

struct _fifo {
   unsigned capacity;
   unsigned head;
   unsigned tail;
   void * data[0];
};

fifo newFifo(int capacity) {
   // The capacity of a circular buffer is one less than one
   // would think: if the user wants a given capacity, the
   // required array is one cell larger.
   capacity++;

   fifo q = (fifo) malloc(sizeof(struct _fifo) + capacity * (sizeof(void *)));
   q->capacity = (unsigned) capacity;
   q->head = 0;
   q->tail = 0;
   return q;
}

int putFifo(fifo q, void * e) {

   if ((q->head+1) % q->capacity == q->tail) // full?
      return -1;
   q->data[q->head] = e;
   q->head = (q->head+1) % q->capacity;
   return 0;
}

typedef void (*printFn) (void*);

int printFifo(fifo q, printFn f) {

   unsigned i;
   for (i = q->tail; i != q->head; i = (i+1) % q->capacity) {
      f(q->data[i]);
   }
   return 0;
}

static void printLong(void * e) {
   // %ld tells printf to print a long integer
   printf("%ld", (long) e);
}

int main() {
   fifo longq;
   longq = newFifo(3);

   printFifo(longq, printLong);
   return 0;
}

My question is this: in function printLong i'm passing void pointer e so does the cast (long) e makes that adress of e? how come i'm getting a value of data and not the address printed?

for example if i do this i'm getting the address printed not the value:

typedef void (*printFn) (void*);

static void printLong (void * e) {
   printf("%ld", (long) e);
}

int printL (void* l, printFn f) {
   f(l);
}

int main() 
{
   long a = 5;
   long* l = &a;

   printf("%ld\n", *l);
   printL(l, printLong);

   return 0;
}

but if i cast

*(long*) e

i'll get the value. How come first code part prints the value not the address? I guess i don't see something so that's why i'm confused. Please help :)

Don't print pointer values like that. Use %p for your format specifier and just send the pointer. If you want to print the long pointed to by a pointer-to-long, then use %ld and dereference the pointer after an appropriate cast, but make sure it holds an address-of-long or you're walking into UB. Is there some specific reason you're not just declaring long* for your parameter type? — WhozCraig
Does this code example come with a prefix, "Don't write code like this"? Can you include the code from "newFifo"? I'm trying to determine if the "void* data[0]" is an overlay hack or if it is cast-avoidance hack. Either way, there's hack-trickery going on that makes me think the book may need an addendum of exothermic oxidization. — kfsone
I really, really hope this code is not taken from a book... "Programming by a would-be engineer"? — Lundin
Are you forced to use this book, or is this your personal choice? If the latter - flush that one in the toilet and pick some reccomended by people who know C textbook. I doubt it will be one of them. — zubergu
This book was recommended to me. I greatly enjoyed reading it. I'll add code for newFifo in main post as requested. — bob murley

tristan tristan · Accepted Answer · 2013-10-24T07:47:42

in function printLong i'm passing void pointer e so does the cast (long) e makes that adress of e? how come i'm getting a value of data and not the address printed?

no. It casts the pointer to a long int. If you want to get the address of e, use &e instead. But e is already the address of a function... i see no point in getting the address of a argument because it is on the stack.

So basically I guess you might have some misunderstanding on functions.A function name is not like a var name:

var_name has a value and you use &var_name to get the address
function_name itself has the value of the address the function
implementation and &function_name would return the same
value(the same address).

UPDATE:

I call f(q->data[i]); to get to printLong (void* e) - so shouldn't e be pointing to q->data[i] ?

In C, function arguments are passed by value. The value of q->data[i] is passed to printLong() so the argument e would have the same value(the same address) as q->data[i]. so e is a copy of q->data[i]

void pointer to function

2 Answers