Number Base in Bash
arithmetic-expressions, bash
Solution
bash's `[` builtin mostly emulates the old standard `[` command (aka `test`, and yes it's really a command), which doesn't know about these newfangled base marks. But bash's arithmetic expressions (`(( ))`) and conditional expressions (`[[ ]]`) do:
$ x=10#08
$ y=10#20
$ echo $((x+y))
28
$ [ $x -lt $y ] && echo yes
-bash: [: 10#08: integer expression expected
$ /bin/[ $x -lt $y ] && echo yes # This uses external test cmd instead of builtin
[: 10#08: bad number
$ [[ $x -lt $y ]] && echo yes
yes
$ ((x<y)) && echo yes
yes
For purely arithmetic tests, `(( ))` is generally easiest to use. But both are bash extensions (i.e. not available in the brand-X shell), so be sure to start your script with `#!/bin/bash`, not `#!/bin/sh`.
Problem
To force numbers to be interpreted in base10, you can prefix with 10#. Specifically 10#08 and 10#09 will be interpreted as valid decimal numbers, and not invalid octal numbers. (I'm taking the output of `date +%S`) However, it seems I then can't use the variable in comparisons: ``` x=10#08 y=10#20 echo $((x+y)) // (returns 28, as expected) while [ $x -lt $y ] do x=$((x++)) done ``` gives me the error ``` -bash: [: 10#08: integer expression expected ``` Is this a bug in bash?