New Perspectives On Html5, Css3, And Javascript, Loose-leaf Version
6th Edition
ISBN: 9781337685764
Author: Patrick M. Carey
Publisher: Cengage Learning
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Chapter 14, Problem 3CP1
Program Plan Intro
To insert a div element with the ID bubblebox in the section element having ID logosection, to display the bubble images in the file named bu_home.html.
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#include
int main (void)
{
int i, *p, count
}
p
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&count;
= 10%;
for (i = 5; i >= 0; i--) {
count++;
(*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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&count;)?
3. [15 pts] Explain in your own words how the [break.
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if expr] command works. When might you
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 14 Solutions
New Perspectives On Html5, Css3, And Javascript, Loose-leaf Version
Ch. 14 - Prob. 1RACh. 14 - Prob. 2RACh. 14 - Prob. 3RACh. 14 - Prob. 4RACh. 14 - Prob. 5RACh. 14 - Prob. 6RACh. 14 - Prob. 7RACh. 14 - Prob. 8RACh. 14 - Prob. 9RACh. 14 - Prob. 10RA
Ch. 14 - Prob. 11RACh. 14 - Prob. 12RACh. 14 - Prob. 13RACh. 14 - Prob. 14RACh. 14 - Prob. 15RACh. 14 - Prob. 1CP1Ch. 14 - Prob. 2CP1Ch. 14 - Prob. 3CP1Ch. 14 - Prob. 4CP1Ch. 14 - Prob. 5CP1Ch. 14 - Prob. 6CP1Ch. 14 - Prob. 7CP1Ch. 14 - Prob. 8CP1Ch. 14 - Prob. 9CP1Ch. 14 - Prob. 10CP1Ch. 14 - Prob. 11CP1Ch. 14 - Prob. 12CP1Ch. 14 - Prob. 13CP1Ch. 14 - Prob. 14CP1Ch. 14 - Prob. 15CP1Ch. 14 - Prob. 16CP1Ch. 14 - Prob. 17CP1Ch. 14 - Prob. 1CP2Ch. 14 - Prob. 2CP2Ch. 14 - Prob. 3CP2Ch. 14 - Prob. 4CP2Ch. 14 - Prob. 5CP2Ch. 14 - Prob. 6CP2Ch. 14 - Prob. 7CP2Ch. 14 - Prob. 8CP2Ch. 14 - Prob. 9CP2Ch. 14 - Prob. 10CP2Ch. 14 - Prob. 11CP2Ch. 14 - Prob. 12CP2Ch. 14 - Prob. 13CP2Ch. 14 - Prob. 14CP2Ch. 14 - Prob. 15CP2Ch. 14 - Prob. 16CP2Ch. 14 - Prob. 17CP2Ch. 14 - Prob. 18CP2Ch. 14 - Prob. 19CP2Ch. 14 - Prob. 20CP2Ch. 14 - Prob. 1CP3Ch. 14 - Prob. 2CP3Ch. 14 - Prob. 3CP3Ch. 14 - Prob. 4CP3Ch. 14 - Prob. 5CP3Ch. 14 - Prob. 6CP3Ch. 14 - Prob. 7CP3Ch. 14 - Prob. 8CP3Ch. 14 - Prob. 9CP3Ch. 14 - Prob. 10CP3Ch. 14 - Prob. 11CP3Ch. 14 - Prob. 12CP3Ch. 14 - Prob. 13CP3Ch. 14 - Prob. 14CP3Ch. 14 - Prob. 15CP3Ch. 14 - Prob. 16CP3Ch. 14 - Prob. 17CP3Ch. 14 - Prob. 18CP3Ch. 14 - Prob. 19CP3Ch. 14 - Prob. 20CP3Ch. 14 - Prob. 21CP3Ch. 14 - Prob. 22CP3Ch. 14 - Prob. 23CP3Ch. 14 - Prob. 24CP3Ch. 14 - Prob. 25CP3Ch. 14 - Prob. 26CP3Ch. 14 - Prob. 27CP3Ch. 14 - Prob. 28CP3Ch. 14 - Prob. 1CP4Ch. 14 - Prob. 2CP4Ch. 14 - Prob. 3CP4Ch. 14 - Prob. 4CP4Ch. 14 - Prob. 5CP4Ch. 14 - Prob. 6CP4Ch. 14 - Prob. 7CP4
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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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