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中编写整数的最优雅和/或简洁的方法是什么？

Answer 1

会这样吗？

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 2

的bitcoin-bash-tools提供的功能{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 
} 

那些工作与上面的文件中直接定义的字段dcr和base58。