How to convert decimal number to Base58 in Bash - bash

Question

myNumber=$(date +%s) # big number in decimal
myNumberInB58=$(toBase58 $myNumber)
toBase58() {
# <your answer here>
}
What is the most elegant and/or concise way to encode an integer in Base58?

Answer 1

The bitcoin-bash-tools provide the functions {en,de}codeBase58:
decodeBase58() {
echo -n "$1" | sed -e's/^\(1*\).*/\1/' -e's/1/00/g' | tr -d '\n'
dc -e "$dcr 16o0$(sed 's/./ 58*l&+/g' <<<$1)p" |
while read n; do echo -n ${n/\\/}; done
}
encodeBase58() {
echo -n "$1" | sed -e's/^\(\(00\)*\).*/\1/' -e's/00/1/g' | tr -d '\n'
dc -e "16i ${1^^} [3A ~r d0<x]dsxx +f" |
while read -r n; do echo -n "${base58[n]}"; done
}
Those work with the fields dcr and base58 defined directly above in the file.

Answer 2

Would this do?
a=( {1..9} {A..H} {J..N} {P..Z} {a..k} {m..z} )
toBase58() {
# TODO: check that $1 is a valid number
local nb=$1 b58= fiftyeight=${#a[#]}
while ((nb)); do
b58=${a[nb%fiftyeight]}$b58
((nb/=fiftyeight))
done
printf '%s\n' "$b58"
}

Answer 3

Here's another version:
# Order-Preserving Base58 (OPB58).
# Also supports negative numbers.
int2b58() {
# Omit IOlo
local n="$1" i BASE58=$(echo {0..9} {A..H} {J..N} {P..Z} {a..k} {m..n} {p..z} | tr -d ' ')
((n < 0 )) && printf -- '-' && n=$((-n))
for i in $(echo "obase=58; $n" | bc); do
printf ${BASE58:$(( 10#$i )):1}
done; echo
}