#include using namespace std; int main(){ int a 15; %3D int b 12; %3D int c 5; int pa &a; %3D int * pb &b; int * pc &c; %3D pb pc; %D pa pb; %3D II II cout << *pa « "," << *pb << "," << *pc << endl;
#include using namespace std; int main(){ int a 15; %3D int b 12; %3D int c 5; int pa &a; %3D int * pb &b; int * pc &c; %3D pb pc; %D pa pb; %3D II II cout << *pa « "," << *pb << "," << *pc << endl;
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
What is the correct output of the following example?

Transcribed Image Text:#include <iostream>
using namespace std;
int main(){
int a =
15;
int b
12;
int c
5;
int *
pa
&a;
%3D
int * pb = &b;
%D
int *
pc
&c;
pb
= pc;
pa
pb;
%D
<« *pb << "," << *pc << endl;
II II
cout << *pa <«
Expert Solution

Step 1
Here we take three-pointer types variables pa, pb, and pc which are pointing at the memory locations a, b, and c respectively. Now the value at pa is 15 , at pb = 12 and at pc = 5.
Now, pb = pc so, value at pb = 5.
Now pa = pb so value at pa is also equals to 5. So, pa = pb = pc = 5.
Now, we are printing *pa which the value at the address pa i.e. 5, and similarly with pb and pc.
So, we get the output 5,5,5.
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