Size of a linked list node

Question

I find this thing a little tricky and recursive as to how sizeof operator calculates the size of one node in a linked list. I've the below structure as one node in list for example:

struct ll_core
{
    char c_ll;
    struct ll_core * next;
};

printf("size of struct ll_core = %d\n\n",sizeof(struct ll_core));

it gives me an answer 8. Now how does it decides the size 8, because to add sizes of individual struct element it agains encounter the same struct ll_core. So it's a kind of loop or recursion while calculating size. Please excuse me and let me know if I'm missing any basic thing in my head while thinking like this.

Answer 1

It doesn't need the size of the struct again, since the struct only contains a pointer to the struct, not the struct itself.

The size of "pointer to struct ll_core" has nothing to do with size of struct ll_core, they are two distinct types. One is pointer, the other one isn't.

You can't declare a "truly" recursive data structure, since it would be infinite.

Answer 2

Assuming you are running your code on 32 bit machine, size of pointer will be 4 bytes. As structure will be aligned on word boundary, 3 bytes will be padded and so the size will be 8 bytes.

this structure will actually be like,

struct ll_core{
    char       c_11:
    char const byte[3];  //padded bytes for alignment reasons
    struct ll_core *next;
};