如何在 Bash 中将十进制数转换为 Base58

Question

myNumber=$(date +%s) # big number in decimal
myNumberInB58=$(toBase58 $myNumber)

toBase58() {
  # <your answer here>
}

在Base58中编码整数最优雅and/or的简洁方法是什么？

Answer 1

这样可以吗？

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

Answer 2

bitcoin-bash-tools 提供函数 {en,de}codeBase58:

decodeBase58() {
    echo -n "" | sed -e's/^\(1*\).*//' -e's/1/00/g' | tr -d '\n'
    dc -e "$dcr 16o0$(sed 's/./ 58*l&+/g' <<<)p" |
    while read n; do echo -n ${n/\/}; done
}

encodeBase58() {
    echo -n "" | sed -e's/^\(\(00\)*\).*//' -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
}

那些与直接在上面文件中定义的字段 dcr 和 base58 一起工作。

如何在 Bash 中将十进制数转换为 Base58

How to convert decimal number to Base58 in Bash

bash

base-conversion