Question

9

Increment operator does not work on a variable if variable is set to 0

rated 0 times [ 9] [ 0] / answers: 1 / hits: 2151 / 3 Years ago, fri, august 20, 2021, 6:54:02

Incrementing a variable var works in bash when enclosed in double parentheses like (( var++ )). But I have found that it fails if variable is set to 0 beforehand like var=0.

$ a=0

$ ((a++)) && echo "command succeeded" || echo "command failed"

command failed



$ a=1

$ ((a++)) && echo "command succeeded" || echo "command failed"

command succeeded

Can someone explain this behavior?

Environment:

I am using gnome-terminal on Ubuntu Desktop 18.04.5 LTS.

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tubequ

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answered 3 Years ago errettas · Accepted Answer

With credit from here: https://unix.stackexchange.com/questions/146773/why-bash-increment-n-0n-return-error

The return value of (( expression )) does not indicate an error status, but, from the bash manpage:

((expression))
The expression is evaluated according to the rules described below under ARITHMETIC EVALUATION. If the value of the expression is
non-zero, the return status is 0; otherwise the return status is 1.
This is exactly equivalent to let "expression".

In ((a++))you are doing a post increment. The value of a is 0 so 1 is returned, after that, it is incremented.

Compare

$ unset a

$ ((a++)) ; echo Exitcode: $? a: $a

Exitcode: 1 a: 1

versus

$ unset a

$ ((++a)) ; echo Exitcode: $? a: $a

Exitcode: 0 a: 1

A pre-increment, so a has become 1 and 0 is returned.