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Chapter 17: Problem 28

T \(\quad$$F \quad\) In physical memory, the nodes in a linked list may be scattered around.

Short Answer

Expert verified
Answer: True

Step by step solution

01

Understanding Linked List Structure

A linked list is a data structure that consists of nodes, each containing some data and a reference (or pointer) to the next node in the sequence. The nodes in a linked list are not required to be stored in contiguous memory locations, as each node knows the address of its successor. This flexibility allows linked lists to be easily resized and manipulated despite the potential non-contiguous arrangement of its nodes.
02

Analyzing the Physical Memory Layout

In physical memory, the memory is divided into small blocks called cells. Each cell has a unique address. When a program needs memory, it triggers a memory allocation request, and the operating system assigns a block of cells to fulfill the requested memory size.
03

Evaluating the Statement

Since the nodes of a linked list are connected using pointers, they don't need to be stored in contiguous memory locations. They can be scattered around in the physical memory as the pointer in each node will direct to the address of the next node in the sequence. Thus, the statement "In physical memory, the nodes in a linked list may be scattered around" is true.
04

Conclusion

The correct answer is T (True), as nodes in a linked list may be scattered around in physical memory due to the use of pointers that connect nodes in the sequence.

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Most popular questions from this chapter

What are some of the advantages that linked lists have over arrays?

Find as many mistakes as you can. NumberList::~NumberList() { ListNode *nodePtr, *nextNode; nodePtr = head; while (nodePtr != nullptr) { nextNode = nodePtr->next; nodePtr->next = nullptr; nodePtr = nextNode; } }

Write code that defines an STL list container for holding float values.

How is the end of a linked list usually signified?

Find as many mistakes as you can. void NumberList::appendNode(double num) { ListNode *newNode, *nodePtr; // Allocate a new node & store num newNode = new listNode; newNode->value = num; // If there are no nodes in the list // make newNode the first node. if (!head) head = newNode; else // Otherwise, insert newNode. { // Find the last node in the list. while (nodePtr->next) nodePtr = nodePtr->next; // Insert newNode as the last node. nodePtr->next = newNode; } }
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