PAC MindLink MindTap Enh Discovering Computers & Microsoft Office 2013: Fundamental
13th Edition
ISBN: 9781305492394
Author: Cenage Learning
Publisher: Cengage Learning
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Chapter 6, Problem 29SG
Explanation of Solution
CMOS technology and its uses:
- CMOS means Complementary Metal-Oxide Semiconductor.
- It offers high speed and consumes little power.
- It uses battery power in order to keep information even when the computer power is in off mode.
- This technology has battery-backed CMOS memory chips...
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#include
int main (void)
{
int i, *p, count
}
p
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= 10%;
for (i = 5; i >= 0; i--) {
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(*p) ++;
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printf("count
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=
%d, Have a wonderful day.\n", count);
1. [20 pts] What is the output of the program? Please explain why.
2. [15 pts] What is the gdb command to set a breakpoint in line 6 (p
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3. [15 pts] Explain in your own words how the [break.
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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.
Chapter 6 Solutions
PAC MindLink MindTap Enh Discovering Computers & Microsoft Office 2013: Fundamental
Ch. 6 - Prob. 1SGCh. 6 - Prob. 2SGCh. 6 - Prob. 3SGCh. 6 - Prob. 4SGCh. 6 - Prob. 5SGCh. 6 - Prob. 6SGCh. 6 - Prob. 7SGCh. 6 - Prob. 8SGCh. 6 - Prob. 9SGCh. 6 - Prob. 10SG
Ch. 6 - Prob. 11SGCh. 6 - Prob. 12SGCh. 6 - Prob. 13SGCh. 6 - Prob. 14SGCh. 6 - Prob. 15SGCh. 6 - Prob. 16SGCh. 6 - Prob. 17SGCh. 6 - Prob. 18SGCh. 6 - Prob. 19SGCh. 6 - Prob. 20SGCh. 6 - Prob. 21SGCh. 6 - Prob. 22SGCh. 6 - Prob. 23SGCh. 6 - Prob. 24SGCh. 6 - Prob. 25SGCh. 6 - Prob. 26SGCh. 6 - Prob. 27SGCh. 6 - Prob. 28SGCh. 6 - Prob. 29SGCh. 6 - Prob. 30SGCh. 6 - Prob. 31SGCh. 6 - Prob. 32SGCh. 6 - Prob. 33SGCh. 6 - Prob. 34SGCh. 6 - Prob. 35SGCh. 6 - Prob. 36SGCh. 6 - Prob. 37SGCh. 6 - Prob. 38SGCh. 6 - Describe how bus width and word size affect and...Ch. 6 - Prob. 40SGCh. 6 - Prob. 41SGCh. 6 - Prob. 42SGCh. 6 - Prob. 43SGCh. 6 - Prob. 44SGCh. 6 - Prob. 45SGCh. 6 - Prob. 46SGCh. 6 - Prob. 47SGCh. 6 - Prob. 1TFCh. 6 - Prob. 2TFCh. 6 - Prob. 3TFCh. 6 - Prob. 4TFCh. 6 - Prob. 5TFCh. 6 - Prob. 6TFCh. 6 - Prob. 7TFCh. 6 - Prob. 8TFCh. 6 - Prob. 9TFCh. 6 - Prob. 10TFCh. 6 - Prob. 11TFCh. 6 - Prob. 12TFCh. 6 - Prob. 1MCCh. 6 - Prob. 2MCCh. 6 - Prob. 3MCCh. 6 - Prob. 4MCCh. 6 - Prob. 5MCCh. 6 - Prob. 6MCCh. 6 - Prob. 7MCCh. 6 - Prob. 8MCCh. 6 - Prob. 1MCh. 6 - Prob. 2MCh. 6 - Prob. 3MCh. 6 - Prob. 4MCh. 6 - Prob. 5MCh. 6 - Prob. 6MCh. 6 - Prob. 7MCh. 6 - Prob. 8MCh. 6 - Prob. 9MCh. 6 - Prob. 10MCh. 6 - Prob. 2CTCh. 6 - Prob. 3CTCh. 6 - Prob. 4CTCh. 6 - Prob. 5CTCh. 6 - Prob. 6CTCh. 6 - Prob. 7CTCh. 6 - Prob. 8CTCh. 6 - Prob. 9CTCh. 6 - Prob. 10CTCh. 6 - Prob. 11CTCh. 6 - Prob. 12CTCh. 6 - Prob. 13CTCh. 6 - Prob. 14CTCh. 6 - Prob. 15CTCh. 6 - Prob. 16CTCh. 6 - Prob. 17CTCh. 6 - Prob. 18CTCh. 6 - Prob. 19CTCh. 6 - Prob. 20CTCh. 6 - Prob. 21CTCh. 6 - Prob. 22CTCh. 6 - Prob. 23CTCh. 6 - Prob. 24CTCh. 6 - Prob. 25CTCh. 6 - Prob. 26CTCh. 6 - Prob. 27CTCh. 6 - Prob. 1PSCh. 6 - Prob. 2PSCh. 6 - Prob. 3PSCh. 6 - Prob. 4PSCh. 6 - Prob. 5PSCh. 6 - Prob. 6PSCh. 6 - Prob. 7PSCh. 6 - Prob. 8PSCh. 6 - Prob. 9PSCh. 6 - Prob. 10PSCh. 6 - Prob. 11PSCh. 6 - Prob. 1.1ECh. 6 - Prob. 1.2ECh. 6 - Prob. 1.3ECh. 6 - Prob. 2.1ECh. 6 - Prob. 2.2ECh. 6 - Prob. 2.3ECh. 6 - Prob. 3.1ECh. 6 - Prob. 3.2ECh. 6 - Prob. 3.3ECh. 6 - Prob. 4.1ECh. 6 - Prob. 4.2ECh. 6 - Prob. 4.3ECh. 6 - Prob. 5.1ECh. 6 - Prob. 5.2ECh. 6 - Prob. 5.3ECh. 6 - Prob. 2IRCh. 6 - Prob. 4IRCh. 6 - Prob. 5IRCh. 6 - Prob. 1CTQCh. 6 - Prob. 2CTQCh. 6 - Prob. 3CTQCh. 6 - Prob. 4CTQ
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- #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!arrow_forward(OnlineGDB) 1. Please read and run the following code and answer the questions.#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;} What is the output of the program? Please explain why.arrow_forward1.[30 pts] Answer the following questions: a. [10 pts] Write a Boolean equation in sum-of-products canonical form for the truth table shown below: A B C Y 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 a. [10 pts] Minimize the Boolean equation you obtained in (a). b. [10 pts] Implement, using Logisim, the simplified logic circuit. Include an image of the circuitarrow_forward
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