Sorry for the delay, I went to the store and came back to find allot of very nice posts. After thinking about it, if the first number happens to end in *99,
FIRST_NUMBER='199'
then I would want 299 for my second number and not a repeat of 199.
So I guess this code will do what I need,
SECOND_NUMBER=$((((FIRST_NUMBER+1)/100+1)*100-1))
I plugged this into my script and I am getting the behavior I expect.
Is there some reason why my thought of just replacing the last two chars is ill conceived? The code I posted didn't work and I was getting a substring expression < 0 error.
I guess what I would have done here would have been to test $FIRST_NUMBER,
Code:
if [ "$FIRST_NUMBER" == "$SECOND_NUMBER" ]; then
let "SECOND_NUMBER=$FIRST_NUMBER+100"
fi
to make sure I didn't end up in the same place, but it is nice to do this in one step.
All of the responses so far are assuming that the OLD_NUMBER is non-negative. If OLD_NUMBER can be less than -99, it is a little more complicated, but I think this works:
for OLD_NUMBER in "$@"
do
NEW_NUMBER=$((((OLD_NUMBER<=-100)*(OLD_NUMBER/100*100+1))+((OLD_NUMBER>-100)*((((OLD_NUMBER+1)/100)+1)*100-1))))
printf '%s -> %d\n' "$OLD_NUMBER" "$NEW_NUMBER"
done
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