调用./dice.sh 1 1 1 1
我明白了dc input base must be a number between 2 and 16 inclusive
vim dice.sh
dice=0$(echo $* | tr -d [:space:] | tr 1-6 0-5)
eval "hex=$(dc -e 16o6i${dice}p)"
# Print the last 64 hex digits of the output:
echo $hex | sed -e 's/.*\(.\{64\}\)$/\1/'
为什么会出现这个错误？我想了解为什么会这样。