Can anyone please explain me the output of this C++ pointer code?

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1.00/5 (1 vote)

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C++

#include <iostream>
using namespace std;
int main()
{
 int* a = new int[10];
 for(int i = 0; i<10; i++)
 a[i] = i;
 int* b = &a[2];
 cout << b[3];
}

Output: 5

What I have tried:

I don't understand why the output is 5.

Posted 14-Oct-18 17:35pm

codingBlonde

Updated 14-Oct-18 21:13pm

CPallini

v2

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Jon McKee · Accepted Answer · 2018-10-14T19:29:00

The reason for that output is something called pointer arithmetic. Arrays are stored sequentially in memory. So the offset from one value to the next is the size of the type. Integer = 4 bytes, float = 4 bytes, double = 8 bytes, etc. This is the fundamental reason why pointer arithmetic works. When you say a[2], what you're really saying is "the pointer to the start of the array (i.e. the first element) plus two offsets worth of integers to reach the third value." You can test this by doing "*(a+2)" which will yield the same result as a[2].

So what's happening in the code you posted: the address (&) of a[2] is being assigned to b. Then you're printing to the standard output stream the value of 3 offsets worth of integers (b[3]). This is 5.

b[0] = a[2]
b[1] = a[3]
b[2] = a[4]
b[3] = a[5]