Question

l Grant P. Microsoft PowerPoi.. Mail - elijah joshlin.. Tokyo Ghoul 3 (Sub.. How to Watch The... OS EJ Purpose The purpose of this lab is to learn how to use pointers within an application. Process Use the following main function: include <stdio.h> void main int i; int j: int p i 10 5 20: Add printf statements to display the values assigned to each variable and their addresses. Your output should appear similar to the following: i: 10 3: 20 p: 1244884 Address of i: 1244884 Address of j 1244872 Address of p: 1244860 Notice the difference in the addresses. Next, add the statement to the end of the program: Display the same variables again to see what has changed. Notice that p has been incremented by 4. Next create an array of 10 integers. Using the pointer, p, initialize every element of the array to 0 in a for or while loop and then display the array

using C++

Answer 1

I have used C++ as mentioned in the question , so I have used cout and cin , and header file used is iostream.

Pointer is a variable that stores the address of another variable. Suppose we have a variable int a=10; Let the address of variable a be 0xabcde0 , and let we have a pointer variable int *ptr , so if we do ptr = &a then ptr will be having the value 0xabcde0 , i.e the address of variable "a", we say ptr points to a. This is how pointer works.

Now coming to arrays and pointers :

Suppose we have a array of 5 elements , i.e int arr[5] = {1,2,3,4,5};

Now let we have a pointer variable ptr , i.e int *ptr;

ptr = arr ; // this statement means that now we have assigned base address of array arr to ptr , now the ptr points to the start index of array arr . That is it stores the address of arr[0].

So if we perform a increment like ptr++ or ptr+ 1 , then ptr will point to next index of array. The value will increase by 4 bytes. This is shown below in the code.

So if we do cout<< *(ptr+1) ; // it will print second element of array , i.e arr[1] . Similarly *(ptr+3) means 4^th element of array.

First element of the array will be *ptr , i.e arr[0].

This above concept has been used in the code below.

Your code goes here :

#include<iostream>
using namespace std;
int main(){
int i,j,*p;
i=10;
j=20;
p=&i;
cout<<"i: "<<i<<"\t\t\t\t Address of i : "<<&i<<endl;
cout<<"j: "<<j<<"\t\t\t\t Address of j : "<<&j<<endl;
cout<<"p: "<<p<<"\t\t\t Address of p : "<<&p<<endl;
p++;
cout<<endl;
cout<<"i: "<<i<<"\t\t\t\t Address of i : "<<&i<<endl;
cout<<"j: "<<j<<"\t\t\t\t Address of j : "<<&j<<endl;
cout<<"p: "<<p<<"\t\t\t Address of p after increment : "<<&p<<endl;
cout<<"\nInitializing and printing array using pointer p : "<<endl;
int arr[10]; /*Integer array of 10 elements*/
p=arr; /*assigning base address of array to p */
for(i=0;i<10;i++){
*(p+i)=0; /* Accessing each index of array using pointer and initialize to 0*/
}
for(i=0;i<10;i++){
cout<<*(p+i)<<endl; /*Printing the array values at each index*/
}
return 0;
}

Screenshots :

Output :

In the above output , notice how the address stored in the pointer variable p changed by 4 bytes when we incremented the pointer using p++ . This happens because the size of the int is 4 bytes and by incrementing address stored in p by 1, we actually increased it by 4 bytes.