I was searching for a way to convert hexadecimal via command line and found there is a very easy method echo $((0x63))
.
It's working great but I'm a little confused as to what is happening here.
I know $(...)
is normally a sub-shell, where the contents are evaluated before the outer command.
Is it still a sub-shell in this situation? I'm thinking not as that would mean the sub-shell is just evaluating (0x63)
which isn't a command.
Can someone break down the command for me?