bash substitution call that increments a variable

Question

I'm trying to define a bash function returning an incremented id that I can access directly using bash substitution:

#!/bin/bash

getId() {
        echo "$x"
        x=$((x+1))
}

x=0
echo "id1: $(getId)"
echo "id2: $(getId)"

However the variable is not incremented and I cannot figure out why.

id1: 0
id2: 0

Please, does someone have an explanation for this behaviour?

Each `$()` is a sub-shell so the assignments aren't happening in the parent shell. — Etan Reisner, Apr 16 '15 at 16:59
The answer [here](http://unix.stackexchange.com/a/124904/67134) might be of use here but I'm not sure offhand. — Etan Reisner, Apr 16 '15 at 17:02
I also thought it was a sub-shell call but I cannot stat it using strace. However, the result is clearely the same. Thanks. — user3149355, Apr 16 '15 at 17:24

Cyrus · Answer 1 · 2015-04-16T17:23:37.197

2

getId() {
  echo "$x"
  ((x++))
}
x=0
echo -n "id1: "
getId
echo -n "id2: "
getId

Output:

id1: 0
id2: 1

edited Apr 16 '15 at 17:23

answered Apr 16 '15 at 17:09

Cyrus

Instead of using a global variable, you could pass the variable name: `getId() { (($1++)); }; getId x; getId x` – glenn jackman Apr 16 '15 at 17:31
@user3149355, the return value if a function is only from 0 to 255. Not very useful. – glenn jackman Apr 16 '15 at 17:32

StackMonkey · Answer 2 · 2015-04-16T17:55:25.280

1

There is no easy way I know of to do it in a sub-shell call using the syntax you have (in the echo line).

An alternate would be:

#!/bin/bash

export x=0

incId() {
        #echo "$x"
        (( x += 1))
}

incId
echo "id1: $x"
incId
echo "id2: $x"

But here you need the out-of-the-echo-line incId function call to get the id incremented.

It also starts counting from 1, not 0.

~~Using the let shell command is the better way to do math too.~~

edited Apr 16 '15 at 17:55

answered Apr 16 '15 at 17:10

StackMonkey

Thanks, you confirm that I'm not creazy and cannot do it in a one-line call. (and thanks for the let command) – user3149355 Apr 16 '15 at 17:26
`let` is an obsolete way to do math; `$((...))` is part of the POSIX standard. – chepner Apr 16 '15 at 17:36
Well, `((...))` is not POSIX; it's a bash extension, but for the most part `((x+=1))` is just short for `: $((x+=1))`. – chepner Apr 16 '15 at 17:57
Some ways to increment with bash: `x=0; : $[ x++ ]; ((x++)); x=$((x+1)); let x=x+1; echo $x` – Cyrus Apr 16 '15 at 18:11

score 1 · Answer 3 · answered Apr 16 '15 at 18:30

1

Might as well make it generic:

incr() { (( $1 += ${2:-1} )); }

Examples:

incr x   ; echo $x  # => 1
incr x   ; echo $x  # => 2
incr x  4; echo $x  # => 6
incr x -2; echo $x  # => 4

answered Apr 16 '15 at 18:30

glenn jackman

3 Answers3