MindTap Computing, 1 term (6 months) Printed Access Card for Vermaat/Sebok/Freund/Campbell Frydenberg's Discovering Computers 2018 (MindTap Course List)
MindTap Computing, 1 term (6 months) Printed Access Card for Vermaat/Sebok/Freund/Campbell Frydenberg's Discovering Computers 2018 (MindTap Course List)
18th Edition
ISBN: 9781337285193
Author: Vermaat, Misty E.; Sebok, Susan L.; Freund, Steven M.; Campbell, Jennifer T.; Frydenberg, Mark
Publisher: Cengage Learning
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Chapter 2, Problem 2.3E

Explanation of Solution

Items that can be post for sale on an online auction:

  • The user can post the hardware items line printers, scanners, cameras, and televisions which are pre-owned, and refurbished.
  • The user can post any items that are pre-owned for sale in an online auction.
  • They can also post kitchen hardware items, sports items, electronic items, fashion items, and so on...

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(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.

Chapter 2 Solutions

MindTap Computing, 1 term (6 months) Printed Access Card for Vermaat/Sebok/Freund/Campbell Frydenberg's Discovering Computers 2018 (MindTap Course List)

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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