I was under the impression that it was possible to access data from a sub-node of a linked list or similar structure by using the arrow and dot operators together like so:
sample->left and sample->right are also pointers, so you want:
if (sample->left->num > sample->right->num) {
// do something
}
Since I don't see it mentioned explicitly:
sample->left gives a struct a*, not a struct a, so we're dealing with pointers.
So you still have to use ->.
You can, however, use sample->left->num.
. is for accessing the members of a struct (or union) e.g.
struct S {
int x;
}
S test;
test.x;
-> is a shorter way to write (*pointer_to_struct).struct_member
Use -> for pointers; use . for objects.
In your specific case you want
if (sample->left->num > sample->right->num)
because all of sample, sample->left, and sample->right are pointers.
If you convert any of those pointers in the pointed to object; use . instead
struct a copyright;
copyright = *(sample->right);
// if (sample->left->num > copyright.num)
if (*(sample->left).num > copyright.num)