Number Base in Bash

Question

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?

Answer 1

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.

Answer 2

How about [ $(($x)) -lt $(($y)) ] or [[ $x -lt $y ]] ?

Answer 3

Not a bug. The shell doesn't do arithmetic evaluation in conditional expressions. See the sections Arithmetic Expansion and ARITHMETIC EVALUATION in man bash for details about when arithmetic evaluation is done.