Can we add an array variable to integer? - c++

Question

This is a program to find median of two sorted arrays.
A divide and conquer based efficient solution to find median of two sorted arrays of same size.
#include<bits/stdc++.h>
using namespace std;
int median(int [], int); /* to get median of a sorted array */
/* This function returns median of ar1[] and ar2[].
Assumptions in this function:
Both ar1[] and ar2[] are sorted arrays
Both have n elements */
int getMedian(int ar1[], int ar2[], int n)
{
/* return -1 for invalid input */
if (n <= 0)
return -1;
if (n == 1)
return (ar1[0] + ar2[0])/2;
if (n == 2)
return (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1])) / 2;
int m1 = median(ar1, n); /* get the median of the first array */
int m2 = median(ar2, n); /* get the median of the second array */
/* If medians are equal then return either m1 or m2 */
if (m1 == m2)
return m1;
/* if m1 < m2 then median must exist in ar1[m1....] and
ar2[....m2] */
if (m1 < m2)
{
if (n % 2 == 0)
{
return getMedian(ar1 + n/2 - 1, ar2, n - n/2 +1);
}
return getMedian(ar1 + n/2, ar2, n - n/2);
}
/* if m1 > m2 then median must exist in ar1[....m1] and
ar2[m2...] */
if (n % 2 == 0)
return getMedian(ar2 + n/2 - 1, ar1, n - n/2 + 1);
return getMedian(ar2 + n/2, ar1, n - n/2);
}
/* Function to get median of a sorted array */
int median(int arr[], int n)
{
if (n%2 == 0)
return (arr[n/2] + arr[n/2-1])/2;
else
return arr[n/2];
}
/* Driver program to test above function */
int main()
{
int ar1[] = {1, 2, 3, 6};
int ar2[] = {4, 6, 8, 10};
int n1 = sizeof(ar1)/sizeof(ar1[0]);
int n2 = sizeof(ar2)/sizeof(ar2[0]);
if (n1 == n2)
printf("Median is %d", getMedian(ar1, ar2, n1));
else
printf("Doesn't work for arrays of unequal size");
return 0;
}
My question is how can I add a array variable to an integer. I means whether it is referring to memory when we do like this (ar1 + n/2 - 1) here in the getmedian function call ?

Answer 1

In expressions an array designator is implicitly converted to the pointer to its first element. Adding an integer to a pointer you will get again a pointer. It is so called the pointer arithmetic.
Thus for example if you have an array
int a[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
const size_t N = sizeof( a ) / sizeof( *a );
then expression
a + N / 2
will point to the sixth element of the array.
Here is a demonstrative program
#include <stdio.h>
int main(void)
{
int a[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
const size_t N = sizeof( a ) / sizeof( *a );
for ( size_t i = 0; i < N; i++ ) printf( "%p: %d\n", a + i, *( a + i ) );
return 0;
}
Its output might look like
0xbfd3d9c8: 0
0xbfd3d9cc: 1
0xbfd3d9d0: 2
0xbfd3d9d4: 3
0xbfd3d9d8: 4
0xbfd3d9dc: 5
0xbfd3d9e0: 6
0xbfd3d9e4: 7
0xbfd3d9e8: 8
0xbfd3d9ec: 9
Also a function parameter declared like an array is also adjusted to pointer. So for example these function declarations
int median(int a[100], int n);
int median(int a[10], int n);
int median(int a[], int n);
int median(int *a, int n);
are equivalent and declare the same one function.