bug#22803: bash printf and negative precision

[Billerbeck, Dirk]

Hello,

I don't know if it's really a bug, if I'm just mistaken or this is the right 
address but I just want to give it a try.

I'm using the following bash-builtin printf command:

FUELLLAENGE=8 ; report="REPIBZVAG" ; REPORTFUELLZEICHEN="0" ; printf 
'%.8s%*.*d\n' $report 0 $((FUELLLAENGE - ${#report} )) "$REPORTFUELLZEICHEN"

The maximum length of $report  is 8 characters but I tested some possible error 
situations so in this example it is 9 characters log. I would expect printf to 
cut the $report from REPIBVZAG to REPIBZVA and to NOT print any additional 
trailing zeros because of the precision value of -1. But a trailing zero is 
printed (set -x output):

+ FUELLLAENGE=8
+ report=REPIBZVAG
+ REPORTFUELLZEICHEN=0
+ printf '%.8s%*.*d\n' REPIBZVAG 0 -1 0
REPIBZVA0

Is this really correct? Or shouldn't the negative precision be taken as zero 
and no character should be printed?

The bash version is "GNU bash, version 3.2.51(1)-release 
(x86_64-suse-linux-gnu)". I know it's an older version but I can't change it as 
there are corporate restrictions. The coreutil package is 
"coreutils-8.12-6.25.32.33.1".

Best regards
Dirk Billerbeck