双向链表中listPointer->leftPointer->rightPointer = newNodePtr;作用及多级指针含义问询

Hey there! Let's break this down nice and slow—chained pointer access like this can feel overwhelming at first, but once you unpack each part, it makes total sense.

First: What's with the chained -> operators? #

In languages like C/C++, the -> operator lets you access a member of a struct/object that's being pointed to. When you chain them (like a->b->c), you're basically "drilling down" one layer at a time:

1. Start with listPointer: this is a pointer to a node in your doubly linked list.
2. listPointer->leftPointer: We take that node, and access its leftPointer member—this is another pointer, which points to the previous node in the list (since it's a doubly linked list, each node has a pointer to the node before it and after it).
3. listPointer->leftPointer->rightPointer: Now we take that previous node, and access its rightPointer member—this pointer normally points to the next node in the list (which, in this case, was originally listPointer's node).

What does this line of code actually do? #

This line is modifying the doubly linked list to insert a new node (newNodePtr) into the list. Specifically:

• We're updating the previous node's "next" pointer to point to the new node instead of its old next node (which was listPointer's node).

Let's use a concrete example to make it real:
Suppose your list looks like this before insertion:

... <-> Node A <-> Node B <-> ...

Let's say listPointer points to Node B. Then:

• listPointer->leftPointer is Node A's pointer.
• listPointer->leftPointer->rightPointer was originally pointing to Node B.
• After running listPointer->leftPointer->rightPointer = newNodePtr;, Node A's rightPointer now points to your new node.

This is just one step in inserting a node into a doubly linked list. You'd also need to:

• Set the new node's leftPointer to point to Node A.
• Set the new node's rightPointer to point to Node B.
• Update Node B's leftPointer to point to the new node.

Once all those steps are done, your list will look like this:

... <-> Node A <-> New Node <-> Node B <-> ...

内容的提问来源于stack exchange，提问作者Ensitti