How use pointer of pointers in C functions?

Question

I have an assignment in which I have to create a function to allocate a block with a starting index multiple of an int called "alignment" and a given size.

function prototype is

int block_alloc(void ** ptr_addr, size_t alignment, size_t size)

assuming that I call block alloc in that way:

int a=0;
ptr = &a;
void* ptr_2 = ptr;
ret = block_alloc ( &ptr_2 ,2* sizeof ( void*) , 2* sizeof ( int ) ) ;

I'm doing in block alloc

 ptr_addr = malloc ( sizeof (int) );

but when printing the pointer it return the &a location

I need to malloc in a given offset and make all work, any hint on what I'm doing wrong?

What about address offset in malloc, it confuses me as I can use only c99 and stdlib functions — Stressed Noob, Sep 20 '21 at 14:45
"of an int called "alignment" and `size_t alignment` confict. Is it `int` or `size_t`. Please explain more what `alignment` is for. — chux - Reinstate Monica, Sep 20 '21 at 14:58
" any hint on what I'm doing wrong?" [@John Bode](https://stackoverflow.com/a/69256464/2410359) has one of the issues identified (a key one). — chux - Reinstate Monica, Sep 20 '21 at 14:59

score 0 · Answer 1 · answered Sep 20 '21 at 14:47

As some programmer dude said, you need to change the expression to *ptr_addr = malloc(sizeof(int));, the explanation for it is as follows:

**ptr_addr is a pointer to the pointer that needs to be modified with the allocated space, so reassigning ptr_addr = malloc(sizeof(int)); just modifies your local copy of the pointer to the pointer and creates a leak instead of modifying the pointer that is being pointed to, hence * dereferences the double pointer and allows you to modify the ptr_2 variable.

Also your malloc call is wrong since you're completely disregarding the arguments and allocating the size for a single int only. The correct malloc call would be malloc(size * alignment); assuming alignment is meant to be the initial size of the array (Guessing from "initial index").

score 0 · Answer 2 · answered Sep 20 '21 at 14:48

Remember the basic pattern:

void update( T *ptr ) // for any type T
{
  *ptr = new_T_value; // updates the thing ptr points to
}

void foo( void )
{
  T var;
  update( &var ); // write a new value to var
}

Now let's replace T with a pointer type P *:

void update( P **ptr )
{
  *ptr = new_Pstar_value; // update the thing ptr points to
}

void foo( void )
{
  P *var;
  update( &var ); // write a new value to var
}

In the second case, var has a pointer type P *, so the input parameter has type P **. But we still want to update the thing ptr points to, so we still write *ptr = ....

So in your case, you would write

*ptr_addr = malloc ( sizeof (int) );

in your block_alloc function, although I think they intend you to allocate memory as

*ptr_addr = malloc( size * alignment );

That allocates enough memory to hold size objects, each of which is alignment bytes wide.

How use pointer of pointers in C functions?

2 Answers2