Question

Working with pointers in C++. I need a brief step by step explanation of the code to help me understand why the output is: 5 4 3 2 1

What will be displayed? Explain each step of the code and the related use of pointers.

#include <iostream>

using namespace std;

int* myfun(int*);

int main()

{

   int x[5] = { 1, 2, 3, 4, 5 };

   int i, *p;

   p = myfun(x);

   for (i = 0; i < 5; i++)

        cout << *(p + i) << " ";

   return 0;

}

int* myfun(int* p)

{

   int i;

   for (i = 0; i < 2; i++) {

        int temp = *(p + i);

        *(p + i) = *(p + 4 - i);

        *(p + 4 - i) = temp;

   }

   return p;

}

Answer 1

#include <iostream>

using namespace std;

int* myfun(int*);

int main()

{

   int x[5] = { 1, 2, 3, 4, 5 };

   int i, *p;

   // p starts with the result of myFun
   // p is an integer pointer, using which we can
   // access elements in the array to which it points
   p = myfun(x);

   // below *(p + i) means p[i], where p is an array
   // and i is the index.
   for (i = 0; i < 5; i++)
        cout << *(p + i) << " ";

   return 0;

}

int* myfun(int* p) {

   int i;

   // Below code basically swaps the values from both ends
   // For a 5 element array, it loops 5/2 = 2 time
   // i = 0 and i = 1
   for (i = 0; i < 2; i++) {

        // (p+i) means ith Location from left corner of array
        // (p+4-i) means, ith location from right corner or array

        // Keep value of left location in temporary variable
        int temp = *(p + i);

        // copy the value on left location from right location
        *(p + i) = *(p + 4 - i);

        // update the value of right location from temporary variable
        *(p + 4 - i) = temp;

   }

   // return the start of the array
   return p;
}

Hi. please find the answer above.. i have given comments so that it is very easy for you to understand the flow. In case of any doubts, please ask in comments. If the answer helps you, please upvote. Thanks!