program to find the frequency of characters in a string

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

int main()
{
   char string[100];
   int c = 0, count[26] = {0}, x;

   printf("Enter a string\n");
   gets(string);

   while (string[c] != '\0') {
   /** Considering characters from 'a' to 'z' only and ignoring others. */

      if (string[c] >= 'a' && string[c] <= 'z') {
         x = string[c] - 'a';
         count[x]++;
      }

      c++;
   }

   for (c = 0; c < 26; c++)
         printf("%c occurs %d times in the string.\n", c + 'a', count[c]);

   return 0;
}

Explanation of "count[string[c]-'a']++", suppose input string begins with 'a' so c is 0 initially and string[0] = 'a' and string[0] - 'a' = 0 and we increment count[0], i.e., 'a' has occurred one time and repeat this until the complete string is scanned.

Output of program:

Did you notice that the string in the output of the program contains every alphabet at least once?

Calculating frequency using function

We will make a function which computes frequency of characters in input string and print it in a table (see output image below).

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

void find_frequency(char [], int []);

int main()
{
   char string[100];
   int c, count[26] = {0};

   printf("Input a string\n");
   gets(string);

   find_frequency(string, count);
 
   printf("Character Count\n");
 
   for (c = 0 ; c < 26 ; c++)
      printf("%c \t  %d\n", c + 'a', count[c]);

   return 0;
}

void find_frequency(char s[], int count[]) {
   int c = 0;
 
   while (s[c] != '\0') {
      if (s[c] >= 'a' && s[c] <= 'z' )
         count[s[c]-'a']++;
      c++;
   }
}

Output of the program:

We do not pass no of elements of array count[] to function find_frequency as they will always be 26 if considering only lower case alphabets. But you can pass if you wish to do so.