Re: retrun value of buildin (( )) / GNU bash, version 4.2.24(1)-release (x86_64-pc-linux-gnu)

[Roman Rakus]

On 07/10/2012 05:32 PM, Orlob Martin (EXT) wrote:
> Hallo,
>
> I was using the build in (( )) function for  count up an index. I think I found 
> a bug in GNU bash, version 4.2.24(1)-release (x86_64-pc-linux-gnu).
((expression)) is one of Compound Commands (not builtin nor function).
Quoting the manual page:
((expression))
The expression is evaluated according to the rules described
below under ARITHMETIC EVALUATION. If the value of the expres‐
sion is non-zero, the return status is 0; otherwise the return
status is 1. This is exactly equivalent to let "expression".

See also a comment below inline.
> The (( )) function is working fine, but the corresponding return value does not 
> match.
>   Check following small example (have the focus on $?)
>
> ----------SCRIPT START-----------------
> #!/bin/bash
>
> index=-3
> for a in 1 2 3 4 5 6 7  ; do
>          echo "Index before: $index"
>          (( index++ ))
>        echo "Return value: $?"
>        echo "Index after: $index"
>        echo "----"
> done
> ----------SCRIPT END------------------
>
> OUTPUT is:
>
>
> Index before: -3
> Return value: 0
> Index after: -2
> ----
> Index before: -2
> Return value: 0
> Index after: -1
> ----
> Index before: -1
> Return value: 0
> Index after: 0
> ----
> Index before: 0
> Return value: 1<-- ?BUG? The index was counted up correctly, from 0 to 1, but return 
> value of (( ))-function is "1"
No, works like intended. index value is 0 (zero) and thus return value 
of (( )) is 1. And the value of index is post-incremented.
> Index after: 1
> ----
> Index before: 1
> Return value: 0
> Index after: 2
> ----
> Index before: 2
> Return value: 0
> Index after: 3
> ----
> Index before: 3
> Return value: 0
> Index after: 4
>
>
> Cheers,
>
> Martin
>
>
>
>
>
>
>
> ----------------------------------------------------------
> Martin Orlob - Systemintegrator
>
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