c - Figuring out the function of cascade of ternary operators? -

i'm trying implement simple (and easy understand) version of merge sort , after bit of research i found following c version:

#include <stdio.h> #include <stdlib.h>  void merge (int *a, int n, int m)  {     //...     int i, j, k;      // allocate memory on heap temp array     int *x = malloc(n * sizeof (int));      // happening here (sorting)     (i = 0, j = m, k = 0; k < n; k++)      {         x[k] = j == n      ? a[i++]              : == m      ? a[j++]              : a[j] < a[i] ? a[j++]              :               a[i++];     }      // deep copy of temp elements initial array     (i = 0; < n; i++)      {         a[i] = x[i];     }      // free temp array     free(x); }  //--------------------------------------------------------------------- void merge_sort (int *a, int n)  {     // base case (subarray 2 elements)     if (n < 2)     {         return;     }      // find index of middle element     int m = n / 2;      // divide 2     merge_sort(a, m);     merge_sort(a + m, n - m);      // sort , merge     merge(a, n, m); }  //===================================================================== int main () {     int a[] = {4, 65, 2, -31, 0, 99, 2, 83, 782, 1};     int n = sizeof / sizeof a[0];     int i;     (i = 0; < n; i++)     {         printf("%d%s", a[i], == n - 1 ? "\n" : " ");     }      merge_sort(a, n);      (i = 0; < n; i++)     {         printf("%d%s", a[i], == n - 1 ? "\n" : " ");     }      return 0; }  

could me describe , comment (possibly break if statements) long cascade of ternary operators in for loop in function merge?

_{content available under gnu free documentation license 1.2 unless otherwise noted.}

what didn't catch has nothing specific merge sort:

    x[k] = j == n      ? a[i++]          : == m      ? a[j++]          : a[j] < a[i] ? a[j++]          :               a[i++]; 

has been written prefer sexy/funny looking code on readiness. real code have lines. could/should expressed as:

if(j == n) {     x[k] = a[i++]; } else if(i == m) {     x[k] = a[j++]; } else if(a[j] < a[i]) {     x[k] = a[j++]; } else {     x[k] = a[i++]; } 

if need understand how merge sort works see merge sort on wikipedia:

---

i've modified bit code old fashion debugging, , used numbers of gif can follow, first here's output:

pass 1. n:2, m:1 inserting 5 inserting 6  pass 2. n:2, m:1 inserting 1 inserting 3  pass 3. n:4, m:2 inserting 1 inserting 3 inserting 5 inserting 6  pass 4. n:2, m:1 inserting 7 inserting 8  pass 5. n:2, m:1 inserting 2 inserting 4  pass 6. n:4, m:2 inserting 2 inserting 4 inserting 7 inserting 8  pass 7. n:8, m:4 inserting 1 inserting 2 inserting 3 inserting 4 inserting 5 inserting 6 inserting 7 inserting 8 

and now, code:

#include <stdio.h> #include <stdlib.h>  static int pass = 0; void merge (int *a, int n, int m) {     int i, j, k;     int *x = malloc(n * sizeof (int));      printf("\npass %d. n:%d, m:%d\n", ++pass, n, m);      (i = 0, j = m, k = 0; k < n; k++) {         if(j == n)         {             printf("inserting %d\n", a[i]);             x[k] = a[i++];         }         else if(i == m)         {             printf("inserting %d\n", a[j]);             x[k] = a[j++];         }         else if(a[j] < a[i])         {             printf("inserting %d\n", a[j]);             x[k] = a[j++];         }         else         {             printf("inserting %d\n", a[i]);             x[k] = a[i++];         }     }     (i = 0; < n; i++) {         a[i] = x[i];     }     free(x); }  void merge_sort (int *a, int n) {     if (n < 2)         return;     int m = n / 2;     merge_sort(a, m);     merge_sort(a + m, n - m);     merge(a, n, m); }  int main () {     int a[] = {6, 5, 3, 1, 8, 7, 2, 4};     int n = sizeof / sizeof a[0];     int i;     (i = 0; < n; i++)         printf("%d%s", a[i], == n - 1 ? "\n" : " ");     merge_sort(a, n);     (i = 0; < n; i++)         printf("%d%s", a[i], == n - 1 ? "\n" : " ");     return 0; }