Question

This part of K&R (The C book) got me thinking:

From the book:

struct tnode {
             char *word;
             int    count;
             struct tnode *left;
             struct tnode *right;
};

The recursion declaration of a node might look chancy, but it’s correct.

Because tnode’s defiition doesn’t use a tnode, but merely a pointer to a tnode somehow the compiler gives us a free pass. But I’m wondering how the computer knows how much memory to give a tnode, when it is declared?

Answer 1

Pointers have a fixed size (32/64 bit depending on the platform) so the compiler knows how much memory is need for the left and right pointers and can calculate the whole size of the struct.

For the same reason if you need a pointer it’s enough to do a forward declaration struct tnode; and you can use a pointer for that struct, eg: struct tree { struct tnode* root; };