[solved] Alphabet

perl -e 'print ++($_="a")'

alphabet="abcdefghijklmnopqrstuvwxyz";
read buchstabe;
read zahl;
str="${alphabet%%$buchstabe*}";
str="${#str}";
str=$[($str + $zahl) % ${#alphabet}];
str="${alphabet:$str:1}";
echo "Zielbuchstabe ist $str";

buchstabe=c
zahl=2
printf "\\$(printf %o $(($(echo $buchstabe | od -An -N1 -tu1) + $zahl)))\n"

buchstabe=a
zahl=1
echo -e "\\0$((((26#$buchstabe+87+$zahl)/64)*100+((26#$buchstabe+87+$zahl)/8%8)*10+((26#$buchstabe+87+$zahl)%8)))"

pascii()
{
printf %b "\\0$(printf %o $(($1%256)))"
}

pascii 0107
pascii 71
pascii 0x47

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

/* printasc.c
   Compile with: gcc -Wall -pedantic -o printasc printasc.c */

int checkDigits(char *a);

int main(int argc, char *argv[])
{
    int a;

    if(argc == 1 || argc > 2)
    {
        puts("Not enough or too many arguments.");
        return 1;
    }

    if (checkDigits(argv[1]) != 0)
    {
        puts("Argument must be a number.");
        return 2;
    }

    a = atoi(argv[1]);

    if (a < -255 || a > 255)
    {
        puts("The argument is either too big or too small.");
        return 3;
    }

    printf("%c\n", a);

    return 0;
}


int checkDigits(char *a)
{
    int i;
    for(i = 0; i < strlen(a); i++)
    {
        if (*(a + i) < '0' || *(a + i) > '9')
        {
            if (*a != '-')
            {
                return 1;
            }
        }
    }
    return 0;
}

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[])
{
    char *err;
    unsigned int a;

    if(argc != 2) {
        puts("Not enough or too many arguments.");
        return 1;
    }

    a = strtoul(argv[1], &err, 0);
    if (err != NULL && *err != '\0') {
        puts("Argument must be a number.");
        return 2;
    }

    if (a > 255) {
        puts("The argument is either too big or too small.");
        return 3;
    }

    printf("%c\n", a);
    return 0;
}