Thread: tokenizing alphabetically

The problem that I see is that your *b[] has "ragged" amounts of char space, for each word.

So b[0] has 5 chars (4 + end of string)
but b[2] only has 2 chars in it.

So when you start sorting, you're toast. b[2] has no room for more than 2 chars in it. You can test this for yourself. Put equal length strings in all of *b[]. The sort works great now. Once you start varying the length of the strings, it's goofed.

Could I interest you in a little index sort, with the same sorter? It would leave everything in place, but it can appear (and be accessed in), sorted order.

Bit of magic there, if you like.

Enjoy! Works fine.

Code:

#include <stdio.h>
#include <string.h>

int main() {
  int i, j, temp_num; 
  int index[6];
  char temp[20];
  char *str[6] = {
  {"This"}, 
  {"is"}, 
  {"a"}, 
  {"string"}, 
  {"to"}, 
  {"parse"}
  };       // scanf("%s", str); 
  for(i = 0; i < 6; i++) {
    printf("\n%s %d", str[i], strlen(str[i])); 
  }
  //initialize the index array
  for(i = 0; i < 6; i++)
    index[i] = i;
  
  printf("\n\n");
  for(i = 0; i < 6; i++) {
    for(j = i+1; j < 6; j++) {

                     //normal string sorting:
                     //if((strcmp(str[i], str[j])) > 0) {
                        //strcpy(temp, str[i]);
                        //strcpy(str[i], str[j]);
                        //strcpy(str[j], temp);

      //sorting by index:
      if((strcmp(str[index[i]], str[index[j]])) > 0) {
        temp_num = index[i];
        index[i] = index[j];
        index[j] = temp_num;
      }
    }
  }

  printf("\n\n In Sorted Order:");
  for(i = 0; i < 6; i++)  //print using the index
    printf("\n%s", str[index[i]]);

  printf("\n\n Order in the array is unchangedl: ");
  for(i = 0; i < 6; i++) {
    printf("\n%s %d", str[i], strlen(str[i])); 

       
  printf("\n\n\t\t\t     press enter when ready");

  i = getchar();
  return 0;
}

You would have to use a regular 2D array: b[6][7], or allocate the memory.