[SOLVED] Bash: perform math on negative integers with leading zeros

Issue

Given a math operation in bash like

a=$(( ${b} + 1 ))

how do I handle cases where b can be either positive or negative with leading zeros? If I try

a=$(( 10#${b} + 1 ))

for b=009, this is successful. But if I try the same for b=-001, this returns an error:

10#: invalid integer constant (error token is "10#")

So is there a way to handle these sorts of cases in a single line? Or is there a way to check if b is less than 0? For the latter option, I have tried

[[ $b < 0 ]] && a=$(( ${b} + 1 )) || a=$(( 10#${b} + 1 ))

but this runs into a similar error in the conditional if b is greater than or equal to 008.

PS: the reason I need the 10# prefix is that numbers starting with leading zeros are interpreted as octal otherwise.

Solution

The syntax to specify the base expects the negative sign, if any, before the base specifier.

$ echo $(( -10#0123 + 1 ))
-122

So something like this could work:

b=-009
b_with_base="10#$b"
echo $(( ${b_with_base/10#-/-10#} + 1 ))

outputs -8 with b=-009 and 10 with b=009.

Answered By - joanis

Answer Checked By - Timothy Miller (WPSolving Admin)