PAC MindLink MindTap Enh Discovering Computers & Microsoft Office 2013: Fundamental
13th Edition
ISBN: 9781305492394
Author: Cenage Learning
Publisher: Cengage Learning
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Expert Solution & Answer
Chapter 2, Problem 30SG
Explanation of Solution
Justification:
“No”, it is not ethical to use a fake name online.
Reason:
- There have been many arguments about using a fake name being ethical or unethical...
Explanation of Solution
Protection against identity theft:
- It is always advised not to tap or click links in or reply to spam.
- It is recommended to install a personal firewall and clear or disable web cookies.
- Turning off the file and printer sharing on internet connection and setting up a free email account to use for merchant forms.
- Emails should be filtered by signing up for it...
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Exercise 1 Function and Structure [30 pts]
Please debug the following program and answer the following questions. There is a cycle in a linked
list if some node in the list can be reached again by continuously following the next pointer.
#include
typedef struct node {
int value;
struct node *next;
} node;
int 11_has_cycle (node *first)
if (first
==
node *head =
{
NULL) return 0;
first;
while (head->next != NULL) {
}
if (head
first) {
return 1; }
head = head->next;
return 0;
void test ll_has_cycle () {
int i;
node nodes [6];
for (i = 0; i < 6; i++) {
nodes [i] .next = NULL;
nodes [i].value = i;
}
nodes [0] .next
=
&nodes [1];
nodes [1] .next
=
&nodes [2];
nodes [2] .next
=
&nodes [3];
nodes [3] .next
nodes [4] .next
&nodes [4];
NULL;
nodes [5] .next = &nodes [0];
printf("1. Checking first list for cycles. \n Function 11_has_cycle says it
has s cycle\n\n", 11_has_cycle (&nodes [0])?"a":"no");
printf("2. Checking length-zero list for cycles. \n Function 11_has_cycle
says it has %s…
Chapter 2 Solutions
PAC MindLink MindTap Enh Discovering Computers & Microsoft Office 2013: Fundamental
Ch. 2 - Prob. 1SGCh. 2 - Prob. 2SGCh. 2 - Prob. 3SGCh. 2 - Prob. 4SGCh. 2 - Prob. 5SGCh. 2 - Prob. 6SGCh. 2 - Prob. 7SGCh. 2 - Prob. 8SGCh. 2 - Prob. 9SGCh. 2 - Prob. 10SG
Ch. 2 - Prob. 11SGCh. 2 - Prob. 12SGCh. 2 - Prob. 13SGCh. 2 - Prob. 14SGCh. 2 - Prob. 15SGCh. 2 - Prob. 16SGCh. 2 - Prob. 17SGCh. 2 - Prob. 18SGCh. 2 - Prob. 19SGCh. 2 - Prob. 20SGCh. 2 - Prob. 21SGCh. 2 - Prob. 22SGCh. 2 - Prob. 23SGCh. 2 - Prob. 24SGCh. 2 - Prob. 25SGCh. 2 - Prob. 26SGCh. 2 - Prob. 27SGCh. 2 - Prob. 28SGCh. 2 - Prob. 29SGCh. 2 - Prob. 30SGCh. 2 - Prob. 31SGCh. 2 - Prob. 32SGCh. 2 - Prob. 33SGCh. 2 - Prob. 34SGCh. 2 - Prob. 35SGCh. 2 - Prob. 36SGCh. 2 - Prob. 37SGCh. 2 - Prob. 38SGCh. 2 - Prob. 39SGCh. 2 - Prob. 40SGCh. 2 - Prob. 41SGCh. 2 - Prob. 42SGCh. 2 - Prob. 43SGCh. 2 - Prob. 44SGCh. 2 - Prob. 45SGCh. 2 - Prob. 46SGCh. 2 - Prob. 47SGCh. 2 - Prob. 48SGCh. 2 - Prob. 49SGCh. 2 - Prob. 1TFCh. 2 - Prob. 2TFCh. 2 - Prob. 3TFCh. 2 - Prob. 4TFCh. 2 - Prob. 5TFCh. 2 - Prob. 6TFCh. 2 - Prob. 7TFCh. 2 - Prob. 8TFCh. 2 - Prob. 9TFCh. 2 - Prob. 10TFCh. 2 - Prob. 11TFCh. 2 - Prob. 12TFCh. 2 - Prob. 1MCCh. 2 - Prob. 2MCCh. 2 - Prob. 3MCCh. 2 - Prob. 4MCCh. 2 - Prob. 5MCCh. 2 - Prob. 6MCCh. 2 - Prob. 7MCCh. 2 - Prob. 8MCCh. 2 - Prob. 1MCh. 2 - Prob. 2MCh. 2 - Prob. 3MCh. 2 - Prob. 4MCh. 2 - Prob. 5MCh. 2 - Prob. 6MCh. 2 - Prob. 7MCh. 2 - Prob. 8MCh. 2 - Prob. 9MCh. 2 - Prob. 10MCh. 2 - Prob. 2CTCh. 2 - Prob. 3CTCh. 2 - Prob. 4CTCh. 2 - Prob. 5CTCh. 2 - Prob. 6CTCh. 2 - Prob. 7CTCh. 2 - Prob. 8CTCh. 2 - Prob. 9CTCh. 2 - Prob. 10CTCh. 2 - Prob. 11CTCh. 2 - Prob. 12CTCh. 2 - Prob. 13CTCh. 2 - Prob. 14CTCh. 2 - Prob. 15CTCh. 2 - Prob. 16CTCh. 2 - Prob. 17CTCh. 2 - Prob. 18CTCh. 2 - Prob. 19CTCh. 2 - Prob. 20CTCh. 2 - Prob. 21CTCh. 2 - Prob. 22CTCh. 2 - Prob. 23CTCh. 2 - Prob. 24CTCh. 2 - Prob. 25CTCh. 2 - Prob. 26CTCh. 2 - Prob. 27CTCh. 2 - Prob. 1PSCh. 2 - Prob. 2PSCh. 2 - Prob. 3PSCh. 2 - Prob. 4PSCh. 2 - Prob. 5PSCh. 2 - Prob. 6PSCh. 2 - Prob. 7PSCh. 2 - Prob. 8PSCh. 2 - Prob. 9PSCh. 2 - Prob. 10PSCh. 2 - Prob. 11PSCh. 2 - Prob. 1.1ECh. 2 - Prob. 1.2ECh. 2 - Prob. 1.3ECh. 2 - Prob. 2.1ECh. 2 - Prob. 2.2ECh. 2 - Prob. 2.3ECh. 2 - Prob. 3.1ECh. 2 - Prob. 3.2ECh. 2 - Prob. 3.3ECh. 2 - Prob. 4.1ECh. 2 - Prob. 4.2ECh. 2 - Prob. 4.3ECh. 2 - Prob. 5.1ECh. 2 - Prob. 5.2ECh. 2 - Prob. 5.3ECh. 2 - Prob. 1IRCh. 2 - Prob. 2IRCh. 2 - Prob. 3IRCh. 2 - Prob. 4IRCh. 2 - Prob. 1CTQCh. 2 - Prob. 2CTQCh. 2 - Prob. 3CTQ
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