Thread: Finding a common suffix

I am a new C programmer, so I imagine my errors are fairly obvious. I am trying to use this program to find a common suffix between two words, starting from the end (destination and procrastination would return stination, for example). My method of adding characters to the suffix array is apparently wrong.

What is my error[s]?

I believe line 31, in particular, has something wrong.

Code:

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

int
main(void)
{
  char word_one[80] = "";
  char word_two[80] = "";
  char suffix[80] = "";
  int shorter_length;  

  printf("Word #1: ");
  fgets(word_one, 80, stdin);
  printf("Word #2: ");
  fgets(word_two, 80, stdin);

  if(strlen(word_one) < strlen(word_two)) //if word one is shorter
  {
    shorter_length = strlen(word_one);
  }//if
  else //otherwise...
  {
    shorter_length = strlen(word_two);
  }//else

  for(int x = shorter_length; x > 0; x--) //process strings
  {
    if(word_one[x] == word_two[x]) 
    {
      printf("match found.  %c matches with %c", word_one[x], word_two[x]);
      suffix[x] = word_one[x];  
    }//if  
  }//for

  printf("Common suffix: %s", suffix);

  return 0;
}//int main(void)

Take a piece of paper and draw out step for step this loop, you will find your problem quickly.

Code:

 for(int x = shorter_length; x > 0; x--) //process strings
  {
    if(word_one[x] == word_two[x])
    {
      printf("match found.  %c matches with %c", word_one[x], word_two[x]);
      suffix[x] = word_one[x]; 
    }//if 
  }

Your algorithm needs tweaking. Words with different lengths can't start comparing letters, based on the length of either the shorter, or the longer word.

You will need two variables - one for each word. Each should be set to the length of it's own word.

Code:

len1 = strlen(word1);
len2 = strlen(word2);

Now a while loop makes a pretty intuitive loop

Code:

while(len1 > 0 && len2 > 0 && word[len1] == word[len2]) {
   //rest of your code in here
   //decrement len1 and len2
}

Make sense?

I'd write a function that returns the length of the common suffix, zero if none:

Code:

size_t common_suffix(const char *const string1, const char *const string2)
{
    const size_t  length1 = (string1 != NULL) ? strlen(string1) : 0;
    const size_t  length2 = (string2 != NULL) ? strlen(string2) : 0;
    const size_t  n = (length1 < length2) ? length1 : length2;
    size_t        i = 1; /* Offset before end of string, 1 .. n, inclusive. */

    /* As long as the i'th characters from the end match, increase i. */
    while (i <= n && string1[length1 - i] == string2[length2 - i])
        i++;

    /* The length of the common suffix is i - 1 characters. */
    return i - 1;
}

I used the ternary operator condition ? value-if-true : value-if-false to treat NULL pointers as if they were empty strings, and to save the smaller of the two string lengths into n.

The index i is always between 1 and shorter string length, inclusive, so you can use string[ length - i ] to access the i'th character from the end of the string.

Using your example strings, you can get the common suffix and its length using

Code:

static const char *const one = "destination";
static const char *const two = "procrastination";
size_t common;

common = common_suffix(one, two);
if (common > 0)
    printf("'%s' and '%s' have a %d-character common suffix, '%s'.\n", one, two, (int)common, one + common);
else
    printf("'%s' and '%s' do not have a common suffix.\n", one, two);