Bit field

isogram.h

#ifndef ISOGRAM_H
#define ISOGRAM_H

#include <stdbool.h>

bool is_isogram(const char phrase[]);

#endif

isogram.c

#include "isogram.h"
#include <stdint.h>

bool is_isogram(const char phrase[])
{
   if (!phrase)
      return false;

   uint32_t letter_flags = 0;
   char sub = 'a';

   while (*phrase) {
      char letter = *phrase;
      if (letter >= 'a' && letter <= 'z')
         sub = 'a';
      else if (letter >= 'A' && letter <= 'Z')
         sub = 'A';
      else
         sub = 'X';

      if (sub != 'X') {
         if ((letter_flags & (1 << (letter - sub))) != 0)
            return false;
         else
            letter_flags |= (1 << (letter - sub));
      }
      phrase++;
   }
   return true;
}

This approach uses the ASCII value of the letter to set the corresponding bit position.

The approach starts by checking if the pointer to the input phrase is NULL. If so, then the function returns false.

The uint32_t value for the bit field is initialized with 0. A character to subtract by is initialized to 'a'.

The input phrase is looped through its characters, looking for a character being a through z or A through Z.

• The ASCII value for a is 97, and for z is 122.
• The ASCII value for A is 65, and for Z is 90.

The character to be subtracted by is set according to whether the character is a lowercase letter, uppercase letter, or something else. If a character is neither an uppercase or lowercase letter, it is ignored. Otherwise, the ASCII value of the letter is used to calculate its position in the bit field.

• If a lowercase letter is subtracted by a, then a will result in 0, because 97 minus 97 equals 0. z would result in 25, because 122 minus 97 equals 25. So a would have 1 shifted left 0 places (so not shifted at all) and z would have 1 shifted left 25 places.
• If an uppercase letter is subtracted by A, then A will result in 0, because 65 minus 65 equals 0. Z would result in 25, because 90 minus 65 equals 25. So A would have 1 shifted left 0 places (so not shifted at all) and Z would have 1 shifted left 25 places.

In that way, both a lowercase z and an uppercase Z can share the same position in the bit field.

So, for an unsigned thirty-two bit integer, if the values for a and Z were both set, the bits would look like

      zyxwvutsrqponmlkjihgfedcba
00000010000000000000000000000001

The bitwise AND operator can be used to check if a bit has already been set. If it has been set, the letter is duplicated and false can be returned immediately. If it has not been set, then the bitwise OR operator can be used to set the bit. If the loop completes without finding a duplicate letter (and returning false), the function returns true.