Problem Solving with C++ (10th Edition)
Problem Solving with C++ (10th Edition)
10th Edition
ISBN: 9780134448282
Author: Walter Savitch, Kenrick Mock
Publisher: PEARSON
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Chapter 9.1, Problem 2STE
Program Plan Intro

Pointer in C++:

A pointer is a variable whose value will be another variable’s address. Generally, a pointer variable is declared as follows:

type *var-name;

Here, “type” is the pointer’s base type and “var-name” is the pointer variable name. The asterisk is used to designate a variable as a pointer.

Given code:

The given declaration is as follows.

int* intPtr1, intPtr2;

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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.
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