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)