What does the operator*() do in a linked list iterator?

What does the operator*() do in a linked list iterator?

• A. It returns a reference to the data in the current node.
• B. It advances the iterator to the next node.
• C. It returns a pointer to the next node.
• D. It removes the current node.

Answer: (A)

The dereference operator on a linked-list iterator gives you access to the value stored in the node the iterator is currently pointing to. It returns a reference (or const reference) to that data, so you can read or modify it without moving the iterator. Moving to the next node is done with the increment operator (usually ++), not with dereferencing. If you need a pointer to the next node, other mechanisms or operators (like operator-> or exposing the node) would handle that, but operator* is about accessing the current element. Removing a node isn’t done via dereferencing either; removal is performed through a list operation such as erase.

The dereference operator on a linked-list iterator gives you access to the value stored in the node the iterator is currently pointing to. It returns a reference (or const reference) to that data, so you can read or modify it without moving the iterator. Moving to the next node is done with the increment operator (usually ++), not with dereferencing. If you need a pointer to the next node, other mechanisms or operators (like operator-> or exposing the node) would handle that, but operator* is about accessing the current element. Removing a node isn’t done via dereferencing either; removal is performed through a list operation such as erase.