r...@beelaertsict.nl writes:
> #!/bin/bash
> # test error in while clause
> #
> set -x
> up () {
> i=$(($i+1))
> [ $i -eq 5 ] && j=$(($j+1)) && i=0
> }
> i=0
> j=0
> # while loop should end when i=4 and j=2
> while [ $i -ne 4 -o $j -ne 2 ]
> do
> if [ $i -eq 3 ]
> then
> up && continue
> fi
> echo i=$i j=$j
> up
> done
It would have helped if your example wasn't so very complicated. I
suspect that you are expecting some of the "&&" to act like ";" that is,
the command following will always be executed. If you expect that, you
should write the code to make that explicit. In addition, you didn't
place echo's in places that would be quite informative, especially
around the "continue" and at the bottom of the loop.
If I add some echo's like this:
#!/bin/bash
# test error in while clause
#
set -x
up () {
i=$(($i+1))
[ $i -eq 5 ] && j=$(($j+1)) && i=0
}
i=0
j=0
# while loop should end when i=4 and j=2
while [ $i -ne 4 -o $j -ne 2 ]
do
echo Testing i == 3
if [ $i -eq 3 ]
then
echo Continuing
up && continue
echo continue not executed
fi
echo i=$i j=$j
up
echo At bottom i=$i j=$j
done
and then run it, the relevant segment of the output is
+ echo At bottom i=3 j=2
At bottom i=3 j=2
+ '[' 3 -ne 4 -o 2 -ne 2 ']'
+ echo Testing i == 3
Testing i == 3
+ '[' 3 -eq 3 ']'
+ echo Continuing
Continuing
+ up
+ i=4
+ '[' 4 -eq 5 ']'
+ echo continue not executed
continue not executed
+ echo i=4 j=2
i=4 j=2
+ up
+ i=5
+ '[' 5 -eq 5 ']'
+ j=3
+ i=0
+ echo At bottom i=0 j=3
At bottom i=0 j=3
This makes it clear what is happening: when i == 3, the "if" is entered
and the "up" is executed. But the "up" returns status 1, and so the
"continue" is not executed. Execution continues in the block to the
second "up", which increments i and then rolls over to increment j. The
exit condition is never met.
Dale
