What does this print to the screen? #include <iostream> using namespace std; int main () {...

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What does this print to the screen? #include <iostream> using namespace std; int main () {...

What does this print to the screen?

#include <iostream>

using namespace std;

int main ()

{

int *a;

a = new int[2];

for(int i = 0; i < 2; i++)

*(a + i) = i;

for(int i = 0; i < 2; i++)

cout << a[i] << " ";

delete [] a;

return 0;

}

I'm getting 0 1

Please step me through the process.

engineering Computer-Science

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Answer 1

Answer #1

First of all we declare a pointer a. Now we allocate it memory equal to an array size of 2.

This means that we can now store integers at (a + 0) and (a + 1) locations.

Now in the first for loop, we initialise this new memory in the following way:-

*(a + i) = i for i = 0, 1

This means that store value of i at the memory location (a + i).

Thus, now (a + 0) points to an integer value 0 and (a + 1) points to an integer value 1.

Now *(a + i) , i.e., value at position (a + i) can also be written as a[i]. Thus the next for loop prints value at a and a + 1 locations.

Thus, a[0] = *(a + 0) = 0 and a[1] = *(a + 1) = 1

Thus the output is:-

0 1

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Answer 2

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