Odtnhj

I have two structs. I am trying the make an array of 'struct bird' inside another struct 'struct nest'.

I am having a hard time allocating the correct amount of space for the bird array when I am creating the nest struct.

Below is my code.

struct bird 
 int value;
;
typedef struct bird bird;

struct nest 
 int nb_birds;
 bird * * birds; //bird * = points to the bird struct, * birds = Array with size unknown
;
typedef struct nest nest;

nest * create_nest(int nb_birds) 
 nest * n = (nest *) malloc(sizeof(nest));
 n->nb_birds = nb_birds;

 //This is where I am stuck
 ***n->birds = (bird *) malloc(sizeof(bird) * nb_birds);*** 


 int i;
 for(i = 0; i < nb_birds; i++)
 n->birds[i]=NULL;
 return n;

You want to allocate array of nb_birds pointers to bird structure, so size to allocation is nb_birds * sizeof(bird *).

Then you want to store pointer to this array, so cast should be to address of the first element — address of bird *, i.e. bird **.

Hence,

n->birds = (bird **) malloc(sizeof(bird *) * nb_birds);

p.s. If you want to allocate N objects on which ptr points to, you can write, or, at least, think about as

ptr = (typeof(ptr)) malloc(sizeof(*ptr) * N);

Update:

It should be noted that malloc returns void * pointer that compatible with any pointer type without explicit casting. So, quoted program line can be as short as

ptr = malloc(N * sizeof(*ptr));

Some programmers, despite them well informed about this void * property, strongly prefer to use explicit cast in such cases. I'm not one of them, but I account such casts as stylistіc preference (like () for sizeof operator). So I left the casting in code above because OP use it, and I thought it was his choice.

Neverthless it is needed (at least for answer completeness and for further readers) to note that such cast is unnecessary and excessive.
.

_{Thank you Paul Ogilvie and chux for patient notes in the comments.}