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Re: Document bug of 'for' compound command


From: vanou
Subject: Re: Document bug of 'for' compound command
Date: Sat, 19 Aug 2017 02:19:00 +0900
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:52.0) Gecko/20100101 Thunderbird/52.2.1

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 <address@hidden <mailto:address@hidden>> wrote:

    On Fri, Aug 18, 2017 at 6:22 PM, vanou <address@hidden
    <mailto:address@hidden>> 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





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