Объясните работу compareX в функции библиотеки qsort ()

Я искал код ближайшей пары, и я нашел этот, который использовал библиотечную функцию qsort (). Я в принципе не понял, как работает этот параметр сравнения. Объяснение, связанное с этим конкретным кодом, будет более ценно. Благодарю.

#include <iostream>
#include <float.h>
#include <stdlib.h>
#include <math.h>
using namespace std;

// A structure to represent a Point in 2D plane
struct Point
{
int x, y;
};/* Following two functions are needed for library function qsort().
Refer: http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */

// Needed to sort array of points according to X coordinate
int compareX(const void* a, const void* b)
{
Point *p1 = (Point *)a,  *p2 = (Point *)b;
return (p1->x - p2->x);
}
// Needed to sort array of points according to Y coordinate
int compareY(const void* a, const void* b)
{
Point *p1 = (Point *)a,   *p2 = (Point *)b;
return (p1->y - p2->y);
}

// A utility function to find the distance between two points
float dist(Point p1, Point p2)
{
return sqrt( (p1.x - p2.x)*(p1.x - p2.x) +
(p1.y - p2.y)*(p1.y - p2.y)
);
}

// A Brute Force method to return the smallest distance between two points
// in P[] of size n
float bruteForce(Point P[], int n)
{
float min = FLT_MAX;
for (int i = 0; i < n; ++i)
for (int j = i+1; j < n; ++j)
if (dist(P[i], P[j]) < min)
min = dist(P[i], P[j]);
return min;
}

// A utility function to find minimum of two float values
float min(float x, float y)
{
return (x < y)? x : y;
}// A utility function to find the distance beween the closest points of
// strip of given size. All points in strip[] are sorted accordint to
// y coordinate. They all have an upper bound on minimum distance as d.
// Note that this method seems to be a O(n^2) method, but it's a O(n)
// method as the inner loop runs at most 6 times
float stripClosest(Point strip[], int size, float d)
{
float min = d;  // Initialize the minimum distance as d

// Pick all points one by one and try the next points till the difference
// between y coordinates is smaller than d.
// This is a proven fact that this loop runs at most 6 times
for (int i = 0; i < size; ++i)
for (int j = i+1; j < size && (strip[j].y - strip[i].y) < min; ++j)
if (dist(strip[i],strip[j]) < min)
min = dist(strip[i], strip[j]);

return min;
}

// A recursive function to find the smallest distance. The array Px contains
// all points sorted according to x coordinates and Py contains all points
// sorted according to y coordinates
float closestUtil(Point Px[], Point Py[], int n)
{
// If there are 2 or 3 points, then use brute force
if (n <= 3)
return bruteForce(Px, n);

// Find the middle point
int mid = n/2;
Point midPoint = Px[mid];// Divide points in y sorted array around the vertical line.
// Assumption: All x coordinates are distinct.
Point Pyl[mid+1];   // y sorted points on left of vertical line
Point Pyr[n-mid-1];  // y sorted points on right of vertical line
int li = 0, ri = 0;  // indexes of left and right subarrays
for (int i = 0; i < n; i++)
{
if (Py[i].x <= midPoint.x)
Pyl[li++] = Py[i];
else
Pyr[ri++] = Py[i];
}

// Consider the vertical line passing through the middle point
// calculate the smallest distance dl on left of middle point and
// dr on right side
float dl = closestUtil(Px, Pyl, mid);
float dr = closestUtil(Px + mid, Pyr, n-mid);

// Find the smaller of two distances
float d = min(dl, dr);

// Build an array strip[] that contains points close (closer than d)
// to the line passing through the middle point
Point strip[n];
int j = 0;
for (int i = 0; i < n; i++)
if (abs(Py[i].x - midPoint.x) < d)
strip[j] = Py[i], j++;

// Find the closest points in strip.  Return the minimum of d and closest
// distance is strip[]
return min(d, stripClosest(strip, j, d) );
}

// The main functin that finds the smallest distance
// This method mainly uses closestUtil()
float closest(Point P[], int n)
{
Point Px[n];
Point Py[n];
for (int i = 0; i < n; i++)
{
Px[i] = P[i];
Py[i] = P[i];
}

qsort(Px, n, sizeof(Point), compareX);
qsort(Py, n, sizeof(Point), compareY);

// Use recursive function closestUtil() to find the smallest distance
return closestUtil(Px, Py, n);
}

// Driver program to test above functions
int main()
{
Point P[] = {{2, 3}, {12, 30}, {40, 50}, {5, 1}, {12, 10}, {3, 4}};
int n = sizeof(P) / sizeof(P[0]);
cout << "The smallest distance is " << closest(P, n);
return 0;
}