Advanced Exercism • c

Pangram

Challenge Overview

Determine if a sentence is a pangram.

Visit the pangram exercise on Exercism to read the full instructions and download the exercise files.

Dig Deeper

Bit field

pangram.h

#ifndef PANGRAM_H
#define PANGRAM_H

#include <stdbool.h>

bool is_pangram(const char *sentence);

#endif

pangram.c

#include "pangram.h" 

bool is_pangram(const char *sentence)
{
    if (!sentence)
        return false;
    int phrasemask = 0;
    char letter;
    
   while ((letter = *sentence) != '\0') {
      if (letter >= 'a' && letter <= 'z')
         phrasemask |= 1 << (letter - 'a');
      else if (letter >= 'A' && letter <= 'Z')
         phrasemask |= 1 << (letter - 'A');
      sentence++;
   }
     return phrasemask == 67108863;
}

The is_pangram function starts by using the logical NOT operator (!) to check if the pointer to the input is NULL. There is a difference between a NULL pointer and a valid pointer pointing to a null character ('\0').

An int is defined to keep track of the used letters of the English alphabet. Since only the rightmost 26 bits will be used, there is no need to have unsignedness.

A char is defined for each character to be tested in the loop.

The while loop uses an assignment expression for its condition. The expression (letter = *sentence) has the value of letter after it has been been assigned the value of the dereferenced sentence pointer. If the value of letter is a null character ('\0'), then the while loop will not run. Note that, had the sentence pointer not been checked for being NULL, then trying to dereference it (*sentence) if it were a NULL pointer would be undefined behavior. It likely would result in a segmentation fault.

The ASCII value of the letter is compared with a range of ASCII values to check if it is one of the lowercase English letters.

The ASCII value for a is 97, and for z is 122.
If the lowercase letter is subtracted by 97, 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.
The bitwise OR operator is used to set the bit at that position in the phrasemask.

If the letter is not a lowercase English letter, the ASCII value of the letter is compared with a range of ASCII values to check if it is one of the uppercase English letters.

The ASCII value for A is 65, and for Z is 90.
If the uppercase letter is subtracted by 65, 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
The bitwise OR operator is used to set the bit at that position in the phrasemask.

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

So, for a 32-bit integer, if the values for a and Z were both set, the bits would look like

      zyxwvutsrqponmlkjihgfedcba
00000010000000000000000000000001

The loop increments the sentence pointer so it points to the address of the next char. C arguments are passed by value, meaning that the actual pointer to the input sentence is not passed in, but a copy of the address it’s pointing to. So the address sentence points to when passed in can be incremented directly without affecting the original pointer in the calling code.

After the loop completes, the function returns if the phrasemask value is the same value as when all 26 bits are set. The value for all 26 bits being set is 67108863.

Source: Exercism c/pangram