How to get character's position in alphabet in C language?

Question

Is there a quick way to retrieve given character's position in the english alphabet in C?

Something like: int position = get_position('g');

Answer 1

int position = 'g' - 'a' + 1;

In C, char values are convertible to int values and take on their ASCII values. In this case, 'a' is the same as 97 and 'g' is 103. Since the alphabet is contiguous within the ASCII character set, subtracting 'a' from your value gives its relative position. Add 1 if you consider 'a' to be the first (instead of zeroth) position.

Answer 2

You should also probably take into account upper/lower case. In my expereince, counting from 1, is often dangerous because it can lead to off-by-one bugs. As a rule of thumb I always convert to a 1-based index only when interacting with the user, and use 0-based counting internally, to avoid confusion.

int GetPosition(char c)
{
   if (c >= 'a' && c <= 'z') {
      return c - 'a';
   }
   else if (c >= 'A' && c <= 'Z') {
      return c - 'A';
   }
   else  {
      // Indicate that it isn't a letter.
      return -1;
   }
}

Answer 3

This will work with EBCDIC and is case-insensitive:

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

int getpos (char c)
{
    int pos;
    const char * alphabet = "abcdefghijklmnopqrstuvwxyz";
    const char * found;

    c = tolower (c);
    found = strchr (alphabet, c);
    pos = found - alphabet;
    if (!found)
        pos = 0;
    else if (pos == 26)
        pos = 0;
    else
        pos++;
    return pos;
}

int main ()
{
    char tests[] = {'A', '%', 'a', 'z', 'M', 0};
    char * c;
    for (c = tests; *c; c++) {
        printf ("%d\n", *c - 'a' + 1);
        printf ("%d\n", getpos (*c));
    }
    return 0;
}

See http://codepad.org/5u5uO5ZR if you want to run it.