PAC MindLink MindTap Enh Discovering Computers & Microsoft Office 2013: Fundamental
13th Edition
ISBN: 9781305492394
Author: Cenage Learning
Publisher: Cengage Learning
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Question
Chapter 3, Problem 4CTQ
Program Plan Intro
Cloud computing:
- Cloud computing is a type of computing that relies on computer resources sharing rather than the application running on local computing devices such as servers, computers, tablet, and Smart phones.
- In a cloud computing, the cloud is used as a symbol for the internet, it means that the cloud computing is a type of internet-based computing.
- It includes three types of service categories
- Infrastructure as a service (IaaS)
- Software as a service (SaaS)
- Platform as a service (PaaS)
Program Plan Intro
Wearable device:
- Wearable device is a small mobile computing device communicated by connected with Bluetooth.
- It is connected to smart phone using Bluetooth.
- It includes three types namely, tracker, smart watch and smart glass.
- They monitor the activities related to fitness.
- They can be communicated with smartphone using wireless connection.
- They are used for viewing the information of photos and videos.
Explanation of Solution
Advantages of technology in national security:
- Information technology helps in improving the tracking, processing and storing the data or information.
- High mobile combat units help in durability and reliability.
- Advanced technology helps in securing the services and avoiding the unauthorized access.
- It helps in the flexible functioning...
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Please run and debug the following program and answer the questions.
(OnlineGDB)
#include <stdio.h>int main(void) {int a;char *s;int v0 = 4, v1 = 5, v2 = 6, v3 = 1, v4 = 2;printf("Exercise 1:\n====================\n");switch(v0) {case 0: printf("Hello October\n"); break;case 1: printf("Go Kean!\n"); break;case 2: printf("Academic Building Center \n"); break;case 3: printf("UNION \n"); break;case 4: printf("Go ");case 5: printf("Kean! \n");default: printf("Have a great semester! \n"); break;}for(a=5; a<v1; a++) {printf("Kean");}printf("\n");if (v2 == 6) {s = "Go";}else {s = "Hello";}if(v3 != v4) {printf("%s Kean!\n",s);} else {printf("%s Computer Science!\n",s);}return 0;}
Assume the following codes are added between line 36 (}) and line 38 (return 0;) v0>0 ? ++v1, ++v2 : --v3; Please give the values of v0, v1, v2, v3, and v4 after this line and explain the reason. You can test the program to verify your answer if you like.
#include <stdio.h>int main(void) {int a;char *s;int v0 = 4, v1 = 5, v2 = 6, v3 = 1, v4 = 2;printf("Exercise 1:\n====================\n");switch(v0) {case 0: printf("Hello October\n"); break;case 1: printf("Go Kean!\n"); break;case 2: printf("Academic Building Center \n"); break;case 3: printf("UNION \n"); break;case 4: printf("Go ");case 5: printf("Kean! \n");default: printf("Have a great semester! \n"); break;}for(a=5; a<v1; a++) {printf("Kean");}printf("\n");if (v2 == 6) {s = "Go";}else {s = "Hello";}if(v3 != v4) {printf("%s Kean!\n",s);} else {printf("%s Computer Science!\n",s);}return 0;}
Output:
Exercise 1:====================Go Kean! Have a great semester!
Go Kean!
Please only modify the initial value of v0, v1, v2, v3 and v4 to get the following output. Youneed to show your program output (in the screenshot) and submit the code that youmodified.Exercise 1:====================Hello OctoberKeanHello Computer Science!
Chapter 3 Solutions
PAC MindLink MindTap Enh Discovering Computers & Microsoft Office 2013: Fundamental
Ch. 3 - Prob. 1SGCh. 3 - Prob. 2SGCh. 3 - Prob. 3SGCh. 3 - Prob. 4SGCh. 3 - Prob. 5SGCh. 3 - Prob. 6SGCh. 3 - Prob. 7SGCh. 3 - Prob. 8SGCh. 3 - Prob. 9SGCh. 3 - Prob. 10SG
Ch. 3 - Prob. 11SGCh. 3 - Prob. 12SGCh. 3 - Prob. 13SGCh. 3 - Prob. 14SGCh. 3 - Prob. 15SGCh. 3 - Prob. 16SGCh. 3 - Prob. 17SGCh. 3 - Prob. 18SGCh. 3 - Prob. 19SGCh. 3 - Prob. 20SGCh. 3 - Prob. 21SGCh. 3 - Prob. 22SGCh. 3 - Prob. 23SGCh. 3 - Prob. 24SGCh. 3 - Prob. 25SGCh. 3 - Prob. 26SGCh. 3 - Prob. 27SGCh. 3 - Prob. 28SGCh. 3 - Prob. 29SGCh. 3 - Prob. 30SGCh. 3 - Prob. 31SGCh. 3 - Prob. 32SGCh. 3 - Prob. 33SGCh. 3 - Prob. 34SGCh. 3 - Prob. 35SGCh. 3 - Prob. 36SGCh. 3 - Prob. 37SGCh. 3 - Prob. 38SGCh. 3 - Prob. 39SGCh. 3 - Prob. 40SGCh. 3 - Prob. 41SGCh. 3 - Prob. 42SGCh. 3 - Prob. 43SGCh. 3 - Prob. 44SGCh. 3 - Prob. 45SGCh. 3 - Prob. 46SGCh. 3 - Prob. 47SGCh. 3 - Prob. 48SGCh. 3 - Prob. 49SGCh. 3 - Prob. 1TFCh. 3 - Prob. 2TFCh. 3 - Prob. 3TFCh. 3 - Prob. 4TFCh. 3 - Prob. 5TFCh. 3 - Prob. 6TFCh. 3 - Prob. 7TFCh. 3 - Prob. 8TFCh. 3 - Prob. 9TFCh. 3 - Prob. 10TFCh. 3 - Prob. 11TFCh. 3 - Prob. 12TFCh. 3 - Prob. 1MCCh. 3 - Prob. 2MCCh. 3 - Prob. 3MCCh. 3 - Prob. 4MCCh. 3 - Prob. 5MCCh. 3 - Prob. 6MCCh. 3 - Prob. 7MCCh. 3 - Prob. 8MCCh. 3 - Prob. 1MCh. 3 - Prob. 2MCh. 3 - Prob. 3MCh. 3 - Prob. 4MCh. 3 - Prob. 5MCh. 3 - Prob. 6MCh. 3 - Prob. 7MCh. 3 - Prob. 8MCh. 3 - Prob. 9MCh. 3 - Prob. 10MCh. 3 - Prob. 2CTCh. 3 - Prob. 3CTCh. 3 - Prob. 4CTCh. 3 - Prob. 5CTCh. 3 - Prob. 6CTCh. 3 - Prob. 7CTCh. 3 - Prob. 8CTCh. 3 - Prob. 9CTCh. 3 - Prob. 10CTCh. 3 - Prob. 11CTCh. 3 - Prob. 12CTCh. 3 - Prob. 13CTCh. 3 - Prob. 14CTCh. 3 - Prob. 15CTCh. 3 - Prob. 16CTCh. 3 - Prob. 17CTCh. 3 - Prob. 18CTCh. 3 - Prob. 19CTCh. 3 - Prob. 20CTCh. 3 - Prob. 21CTCh. 3 - Prob. 22CTCh. 3 - Prob. 23CTCh. 3 - Prob. 24CTCh. 3 - Prob. 25CTCh. 3 - Prob. 26CTCh. 3 - Prob. 27CTCh. 3 - Prob. 28CTCh. 3 - Prob. 29CTCh. 3 - Prob. 30CTCh. 3 - Prob. 1PSCh. 3 - Prob. 2PSCh. 3 - Prob. 3PSCh. 3 - Prob. 4PSCh. 3 - Prob. 5PSCh. 3 - Prob. 6PSCh. 3 - Prob. 7PSCh. 3 - Prob. 8PSCh. 3 - Prob. 9PSCh. 3 - Prob. 10PSCh. 3 - Prob. 11PSCh. 3 - Prob. 1.1ECh. 3 - Prob. 1.2ECh. 3 - Prob. 2.1ECh. 3 - Prob. 2.2ECh. 3 - Prob. 2.3ECh. 3 - Prob. 3.1ECh. 3 - Prob. 3.2ECh. 3 - Prob. 3.3ECh. 3 - Prob. 4.1ECh. 3 - Prob. 4.2ECh. 3 - Prob. 4.3ECh. 3 - Prob. 5.1ECh. 3 - Prob. 5.2ECh. 3 - Prob. 5.3ECh. 3 - Prob. 1IRCh. 3 - Prob. 2IRCh. 3 - Prob. 3IRCh. 3 - Prob. 4IRCh. 3 - Prob. 1CTQCh. 3 - Prob. 2CTQCh. 3 - Prob. 3CTQCh. 3 - Prob. 4CTQ
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