How to get the last character of a string in a shell?

Question

I have written the following lines to get the last character of a string:

str=$1
i=$((${#str}-1))
echo ${str:$i:1}

It works for abcd/

Answer 1

For portability you can say "${s#"${s%?}"}":

#!/bin/sh
m=bzzzM n=bzzzN
for s in \
    'vv'  'w'   ''    'uu  ' ' uu ' '  uu' / \
    'ab?' 'a?b' '?ab' 'ab??' 'a??b' '??ab' / \
    'cd#' 'c#d' '#cd' 'cd##' 'c##d' '##cd' / \
    'ef%' 'e%f' '%ef' 'ef%%' 'e%%f' '%%ef' / \
    'gh*' 'g*h' '*gh' 'gh**' 'g**h' '**gh' / \
    'ij"' 'i"j' '"ij' "ij'"  "i'j"  "'ij"  / \
    'kl{' 'k{l' '{kl' 'kl{}' 'k{}l' '{}kl' / \
    'mn$' 'm$n' '$mn' 'mn$$' 'm$$n' '$$mn' /
do  case $s in
    (/) printf '\n' ;;
    (*) printf '.%s. ' "${s#"${s%?}"}" ;;
    esac
done

Output:

.v. .w. .. . . . . .u. 
.?. .b. .b. .?. .b. .b. 
.#. .d. .d. .#. .d. .d. 
.%. .f. .f. .%. .f. .f. 
.*. .h. .h. .*. .h. .h. 
.". .j. .j. .'. .j. .j. 
.{. .l. .l. .}. .l. .l. 
.$. .n. .n. .$. .n. .n.