Thanks Pierre Gaston and DJ Mills,
I misunderstood arithmetic expression and now make sense.
Thank you.
On 08/19/2017 01:12 AM, DJ Mills wrote:
It's an arithmetic expression, 1 is true and 0 is false. This is not
the same as the exit status of a command
On Fri, Aug 18, 2017 at 11:59 AM, Pierre Gaston
mailto:pierre.gas...@gmail.com>> wrote:
On Fri, Aug 18, 2017 at 6:22 PM, vanou mailto:v...@star.ocn.ne.jp>> wrote:
> Hello,
>
> I think, there is document bug related to 'for' compound command
in both
> Man page and Info doc.
>
>
> In man page, description of 'for' compound command ...
>
>
>
> * for (( expr1 ; expr2 ; expr3 )) ; do list ; done
> * First, the arithmetic expression expr1 is evaluated according
> * to the rules described below under ARITHMETIC EVALUATION.
> * The arithmetic expression expr2 is then evaluated
repeatedly
> * until it evaluates to zero.
> <-- not zero, but 1
> * Each time expr2 evaluates to a non-zero value,
> <-- not non-zero,but 0
> * list is executed and the arithmetic expression expr3 is
evaluated.
> * If any expression is omitted, it behaves as if it
evaluates to 1.
> <-- not 1, but 0
> * The return value is the exit status of the last command in
list
> * that is executed, or false if any of the expressions is
invalid.
>
>
>
>
> And same document bug in Info documentation of bash.
>
> This bug is seen at bash 4.4.
>
> Thanks,
> Vanou
>
>
> The documentation seems ok, what makes you think the contrary?
eg
for ((;0;)); do echo foo;done # prints nothing
for ((;1;)); do echo foo;done # is infinte
for ((;;)); do echo foo;done # is also infinite
