Converting Letters to Numbers in C

Question

I\'m trying to write a code that would convert letters into numbers. For example A ==> 0 B ==> 1 C ==> 2 and so on. Im thinking of writing 26 if statements. I\'m wondering

Answer 1

This is a way that I feel is better than the switch method, and yet is standards compliant (does not assume ASCII):

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

/* returns -1 if c is not an alphabetic character */
int c_to_n(char c)
{
    int n = -1;
    static const char * const alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    char *p = strchr(alphabet, toupper((unsigned char)c));

    if (p)
    {
        n = p - alphabet;
    }

    return n;
}

Answer 2

I wrote this bit of code for a project, and I was wondering how naive this approach was.

The benefit here is that is seems to be adherent to the standard, and my guess is that the runtime is approx. O(k) where k is the size of the alphabet.

int ctoi(char c)
{
    int index;
    char* alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    c = toupper(c);

    // avoid doing strlen here to juice some efficiency.
    for(index = 0; index != 26; index++)
    {
        if(c == alphabet[index])
        {
            return index;
        }
    }

    return -1;
}

Answer 3

There is a much better way.

In ASCII (www.asciitable.com) you can know the numerical values of these characters.

'A' is 0x41.

So you can simply minus 0x41 from them, to get the numbers. I don't know c very well, but something like:

int num = 'A' - 0x41;

should work.

Answer 4

If you need to deal with upper-case and lower-case then you may want to do something like:

if (letter >= 'A' && letter <= 'Z')
  num = letter - 'A';
else if (letter >= 'a' && letter <= 'z')
  num = letter - 'a';

If you want to display these, then you will want to convert the number into an ascii value by adding a '0' to it:

  asciinumber = num + '0';