Help in median of two sorted arrays of same size

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1

// A Simple Merge based O(n)
// solution to find median of
// two sorted arrays
#include <bits/stdc++.h>
using namespace std;

/* 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)
{
int i = 0; / Current index of
i/p array ar1[] /
int j = 0; / Current index of
i/p array ar2[] */
int count;
int m1 = -1, m2 = -1;

/* Since there are 2n elements, 
median will be average of elements 
at index n-1 and n in the array 
obtained after merging ar1 and ar2 */
for (count = 0; count <= n; count++) 
{ 
	/* Below is to handle case where 
	all elements of ar1[] are 
	smaller than smallest(or first) 
	element of ar2[]*/
	if (i == n) 
	{ 
		m1 = m2; 
		m2 = ar2[0]; 
		break; 
	} 

	/*Below is to handle case where 
	all elements of ar2[] are 
	smaller than smallest(or first) 
	element of ar1[]*/
	else if (j == n) 
	{ 
		m1 = m2; 
		m2 = ar1[0]; 
		break; 
	} 
	/* equals sign because if two 
	arrays have some common elements */
	if (ar1[i] <= ar2[j]) 
	{ 
		/* Store the prev median */
		m1 = m2; 
		m2 = ar1[i]; 
		i++; 
	} 
	else
	{ 
		/* Store the prev median */
		m1 = m2; 
		m2 = ar2[j]; 
		j++; 
	} 
} 

return (m1 + m2)/2;

}

// Driver Code
int main()
{
int ar1[] = {1, 12, 15, 26, 38};
int ar2[] = {2, 13, 17, 30, 45};

int n1 = sizeof(ar1) / sizeof(ar1[0]); 
int n2 = sizeof(ar2) / sizeof(ar2[0]); 
if (n1 == n2) 
	cout << "Median is "
		<< getMedian(ar1, ar2, n1) ; 
else
	cout << "Doesn't work for arrays"
		<< " of unequal size" ; 
getchar(); 
return 0;

}

in this code why they are calling getMedian() with arr1+n/2 when m1 < m2