How to change letter frequency to percentage?

Question

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

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

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

   while ( string[c] != '\0' )
   {

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

      else if (string[c] >= 'A' && string[c] <= 'Z')
         count[string[c]-'A']++;
      c++;

   }

   for ( c = 0 ; c < 26 ; c++ )
   {
      if( count[c] != 0 )
     printf( "%c %d\n", c+'a', count[c]);
   }

   return 0;
}

So I managed to get the code working to count the letter frequencies as a number. But my assignment tells me to represent it as a percentage of the whole string.

So for example, the input aaab would give me a - 0.7500, b - 0.2500.

How would I modify this code to represent it as a percentage rather than a number?

Also, if I were to do this where the user inputs strings until EOF, do I just delete the "Enter a string" print statement and change while ( string[c] != '\0' ) to while ( string[c] != EOF )?

Answer 1

Just add an accumulative variable that each time that it reads a character, it is incremented by one (assuming that you want to count frequency among valid characters a-z and A-Z). In the end, divide the count by such variable.

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

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

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

   while ( string[c] != '\0' )
   {

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

      else if (string[c] >= 'A' && string[c] <= 'Z'){
          count[string[c]-'A']++;
          accum++;
      }
      c++;
   }

   for ( c = 0 ; c < 26 ; c++ )
   {
      if( count[c] != 0 )
          printf( "%c %f\n", c+'a', ((double)count[c])/accum);
   }

   return 0;
}