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 <pierre.gas...@gmail.com> wrote: > On Fri, Aug 18, 2017 at 6:22 PM, vanou <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 >
