Re: Logical operators in arithmetic evaluation: documentation vs implementation

[Pierre Gaston]

On Thu, Apr 18, 2013 at 12:59 PM, andras@coolbox.se <andras@coolbox.se>wrote:

> The ARITHMETIC EVALUATION section of the man page claims equivalence with
> C for all the operators, but in reality bash does not perform short circuit
> evaluation, which implies that the logical operators do NOT produce the
> same results as in C.
> Try these, for example:
>
> f () {
> # echo "$@" >&2
>     local n=$1
>     echo $((0 < n ? n * $(f $((n-1))) : 1))
> }
>
> or
>
> g() {
> # echo "$@" >&2
>     local a=$1 b=$2
>     echo $((0 == b ? a : $(g b $((a%b)))))
> }
>
> Note that && and || are affected the same way, and the side effect is not
> due solely to recursion.
>
> $ echo $((1 || $(echo + >&2 && echo 0)))
>
> $ echo $((0 && $(echo + >&2 && echo 1)))
>
> The results are correct, but the side effects are NOT the same as in C.
>
> This may all be fine and as intended, but in that case the documentation
> feels somewhat misleading.
>
> Best,
>
> --@;
>


Expansions are not part of the arithmetic evaluation and are done before.
So your $( )  and $(( )) inside the outer $(( ))  are done before the
arithmetic evaluation starts.

For the evaluation bash does short-circuit:
$ n=0;echo $(( n?++n:n))
0
$ n=1;echo $(( n?++n:n))
2
$ n=0;echo $(( 1 && ++nn));echo $n
1
0