6

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.

muru
  • 207,228
Saad
  • 111

2 Answers2

10

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.

vanadium
  • 97,564
6

This works for me (in bash in Ubuntu),

$ a=0
$ echo $((a++))
0
$ echo $((a++))
1
$ echo $((a++))
2
$ echo $((a++))
3
$ echo $a
4

Notice the difference with

$ a=0
$ echo $((++a))
1
$ echo $((++a))
2
$ echo $((++a))
3
$ echo $a
3
sudodus
  • 47,684