Re: increment & decrement error when variable is 0

[felix]

On Mon, Nov 23, 2020 at 05:20:22PM -0500, Greg Wooledge wrote:
> On Mon, Nov 23, 2020 at 07:36:29PM +0000, Jetzer, Bill wrote:
>...
> Exercises 1 and 2 apply directly...

>From man bash:
       ((expression))
              The expression is evaluated according to the rules described be‐
              low under ARITHMETIC EVALUATION.  If the value of the expression
              is non-zero, the return status is 0; otherwise the return status
              is 1.  This is exactly equivalent to let "expression".

So little modified test/demo:

    for ((i=-100; i<100; i++)) ;do
        x=$i
        ((v = --x ,v)) || echo "err code $? on --x going from $i to $x -> $v"
        x=$i
        ((v = x-- ,v)) || echo "err code $? on x-- going from $i to $x -> $v"
        x=$i
        ((v = ++x ,v)) || echo "err code $? on ++x going from $i to $x -> $v"
        x=$i
        ((v = x++ ,v)) || echo "err code $? on x++ going from $i to $x -> $v"
    done

Will render near same output:

    err code 1 on ++x going from -1 to 0 -> 0
    err code 1 on x-- going from 0 to -1 -> 0
    err code 1 on x++ going from 0 to 1 -> 0
    err code 1 on --x going from 1 to 0 -> 0

Where *when the value of the expression is zero, the return status is 1*.

Difference between ((x--)) and ((--x)) is only usefull when *using* this
variable simultaneously while decrementing them (for assignment, testing,
or printing):

    i=10;while ((i));do echo $i;((i--));done
and
    i=10;while ((i));do echo $i;((--i));done
will do same result.
But
    i=10;while ((i));do echo $((i--));done  # return 10 to 1
    i=10;while ((i));do echo $((--i));done  # print 9 to 0
or
    i=10;while ((i--));do echo $i;done   # print 9 to 0
    i=10;while ((--i));do echo $i;done   # print 9 to 1 (exit at 0 before print)

-- 
 Félix Hauri  -  <felix@f-hauri.ch>  -  http://www.f-hauri.ch