Question

C++

1. What is the final output?

2. Why is 2 the ouput for (int)*c? Please explain what happens during the line c = c + *p +1;

#include <iostream>
using namespace std;

int main() {
   int arr[5] = { 5,2,3,1,4 };

   int* p = arr;

   cout << (int)*p << endl;

   unsigned char* c = (unsigned char*)arr;

   p = p + 2;

   c = c + *p + 1;
  
   cout << (int)*p << endl;
   cout << (int)*c << endl;

   return 0;

}

Answer 1

Answer 1:

Output is 5,3,2

p is pointing to arr so when we access the P it will print the first element of the array

we are adding p+2 so now it will moves 2 blocks so points to 3rd element which 3 so it prints 3

so now are c is pointing to starting of the array , next we are adding c+*p+1 means c+3+1 so we are adding 4 to c it will move 4 blocks

In 2nd cout we are type casting to int * means now we are converting back to int address so adding 4 to int address means it will point to 2 block so it will print 2

Answer 2: there we are converting it to Int address to get the 2nd element from the array

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