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MindTap Computing for Vermaat's Enhanced Discovering Computers, 1st Edition, [Instant Access], 1 term (6 months)
1st Edition
ISBN: 9781285845937
Author: Misty E. Vermaat
Publisher: Cengage Archive
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Expert Solution & Answer
Chapter 7, Problem 40SG
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#include
int main (void)
{
int i, *p, count
}
p
=
&count;
= 10%;
for (i = 5; i >= 0; i--) {
count++;
(*p) ++;
}
printf("count
return 0;
=
%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
=
&count;)?
3. [15 pts] Explain in your own words how the [break.
need to use such command?
...
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 7 Solutions
MindTap Computing for Vermaat's Enhanced Discovering Computers, 1st Edition, [Instant Access], 1 term (6 months)
Ch. 7 - Prob. 1SGCh. 7 - Prob. 2SGCh. 7 - Prob. 3SGCh. 7 - Prob. 4SGCh. 7 - Prob. 5SGCh. 7 - Prob. 6SGCh. 7 - Prob. 7SGCh. 7 - Prob. 8SGCh. 7 - Prob. 9SGCh. 7 - Prob. 10SG
Ch. 7 - Prob. 12SGCh. 7 - Prob. 13SGCh. 7 - Prob. 14SGCh. 7 - Prob. 15SGCh. 7 - Prob. 16SGCh. 7 - Prob. 17SGCh. 7 - Prob. 18SGCh. 7 - Prob. 19SGCh. 7 - Prob. 20SGCh. 7 - Prob. 21SGCh. 7 - Prob. 22SGCh. 7 - Prob. 23SGCh. 7 - Prob. 24SGCh. 7 - Prob. 25SGCh. 7 - Prob. 26SGCh. 7 - Prob. 27SGCh. 7 - Prob. 28SGCh. 7 - Prob. 29SGCh. 7 - Prob. 30SGCh. 7 - Prob. 31SGCh. 7 - Prob. 32SGCh. 7 - Prob. 33SGCh. 7 - Prob. 34SGCh. 7 - Prob. 35SGCh. 7 - Prob. 36SGCh. 7 - Prob. 37SGCh. 7 - Prob. 38SGCh. 7 - Prob. 39SGCh. 7 - Prob. 40SGCh. 7 - Prob. 41SGCh. 7 - Prob. 42SGCh. 7 - Prob. 43SGCh. 7 - Prob. 44SGCh. 7 - Prob. 45SGCh. 7 - Prob. 46SGCh. 7 - Prob. 47SGCh. 7 - Prob. 48SGCh. 7 - Prob. 49SGCh. 7 - Prob. 1TFCh. 7 - Prob. 2TFCh. 7 - Prob. 3TFCh. 7 - Prob. 4TFCh. 7 - Prob. 5TFCh. 7 - Prob. 6TFCh. 7 - Prob. 7TFCh. 7 - Prob. 8TFCh. 7 - Prob. 9TFCh. 7 - Prob. 10TFCh. 7 - Prob. 11TFCh. 7 - Prob. 12TFCh. 7 - Prob. 2MCCh. 7 - Prob. 3MCCh. 7 - Prob. 4MCCh. 7 - Prob. 5MCCh. 7 - Prob. 6MCCh. 7 - Prob. 7MCCh. 7 - Prob. 8MCCh. 7 - Prob. 1MCh. 7 - Prob. 2MCh. 7 - Prob. 3MCh. 7 - Prob. 4MCh. 7 - Prob. 5MCh. 7 - Prob. 6MCh. 7 - Prob. 7MCh. 7 - Prob. 8MCh. 7 - Prob. 9MCh. 7 - Prob. 10MCh. 7 - Prob. 2CTCh. 7 - Prob. 3CTCh. 7 - Prob. 4CTCh. 7 - Prob. 5CTCh. 7 - Prob. 6CTCh. 7 - Prob. 7CTCh. 7 - Prob. 8CTCh. 7 - Prob. 9CTCh. 7 - Prob. 10CTCh. 7 - Prob. 11CTCh. 7 - Prob. 12CTCh. 7 - Prob. 13CTCh. 7 - Prob. 14CTCh. 7 - Prob. 15CTCh. 7 - Prob. 16CTCh. 7 - Prob. 17CTCh. 7 - Prob. 18CTCh. 7 - Prob. 20CTCh. 7 - Prob. 21CTCh. 7 - Prob. 22CTCh. 7 - Prob. 23CTCh. 7 - Prob. 24CTCh. 7 - Prob. 25CTCh. 7 - Prob. 26CTCh. 7 - Prob. 27CTCh. 7 - Prob. 28CTCh. 7 - Prob. 1PSCh. 7 - Prob. 2PSCh. 7 - Prob. 3PSCh. 7 - Prob. 4PSCh. 7 - Prob. 5PSCh. 7 - Prob. 6PSCh. 7 - Prob. 7PSCh. 7 - Prob. 8PSCh. 7 - Prob. 9PSCh. 7 - Prob. 10PSCh. 7 - Prob. 11PSCh. 7 - Prob. 1.1ECh. 7 - Prob. 1.2ECh. 7 - Prob. 1.3ECh. 7 - Prob. 2.1ECh. 7 - Prob. 2.2ECh. 7 - Prob. 2.3ECh. 7 - Prob. 3.1ECh. 7 - Prob. 3.2ECh. 7 - Prob. 4.1ECh. 7 - Prob. 4.2ECh. 7 - Prob. 4.3ECh. 7 - Prob. 5.1ECh. 7 - Prob. 5.2ECh. 7 - Prob. 5.3ECh. 7 - Prob. 1IRCh. 7 - Prob. 2IRCh. 7 - Prob. 4IRCh. 7 - Prob. 5IRCh. 7 - Prob. 1CTQCh. 7 - Prob. 2CTQCh. 7 - Prob. 3CTQ
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