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I'm trying to pad zero to a number
a=010
printf "%04d" $a
0008
I need the "output" as "0010" it's converting the value.
and even tried with typeset
a=010
typeset -RZ2 a
echo $a
when I use the same in my script I got the following error:
Invalid -R option
Don't put a "0" at the beginning of the number -- it treats the number in the octal base. Simply assign the decimal number.
a=10
printf "%04d" $a
0010
If you are reading the numbers from somewhere else, you may consider removing the 0s at the beginning as follows:
a=0010
b=$(echo $a | sed 's/^0*//')
printf "%04d" $a
0008
printf "%04d" $b
0010
b=$(sed s/^0*// <<<"$a")
Mar 14, 2017 at 6:52
%08.0fAs %f is used to print decimal floating point, argument must be decimal.
So there is no risk to see argument interpreted as octal:
Do simply:
a=0010
a=`printf %08.0f $a`
or
a=$(printf %08.0f $a)
then
echo $a
00000010
This work find under any POSIX shell
Using bash, you could avoid fork by using -v option of printf:
a=0010
printf -v a %08.0f $a
echo $a
00000010
%d)You can drop the left 0 character with the following:
a=00010
while [ "$a" != "${a#0}" ] ;do a=${a#0};done
printf "%08d" $a
00000010
As this doesn't fork a new command session, this could be quicker than using sed.
You can force decimal interpretation with the following syntax:
a=000010
printf "%08d" $((10#$a))
00000010
This could be useful for conversions:
a="0a"
printf "%04d" $((16#$a))
0010
a="00001010"
printf "%04d" $((2#$a))
0010
a="00012"
printf "%04d" $((8#$a))
0010
a="0020"
printf "%04d" $((5#$a))
0010
a="0013"
printf "%04d" $((7#$a))
0010
and so on...
a="zz"
printf "%04d" $((36#$a))
1295
$() generate a new command session (fork), echo $a will run in this new session, than | sed will generate a new command session to pipe to /bin/sed outer binary.
Aug 26, 2013 at 14:40
You can accomplish that by using a different conversion specifier, for example the "f" specifier. From the printf manual:
f, F
The double argument is rounded and converted to decimal notation in the style [-]ddd.ddd, where the number of digits after the decimal-point character is equal to the precision specification. If the precision is missing, it is taken as 6; if the precision is explicitly zero, no decimal-point character appears. If a decimal point appears, at least one digit appears before it.
But we don't really want a floating point representation of your number, so we must specify a precision of zero. This should do it:
a=010
printf "%04.0f" $a
I'm assuming you are using Linux, but this should also work with other flavours of Unix.
man printf, u could use: printf "%04g" 010! +1, interesting way to explore too...
Aug 26, 2013 at 14:51
typeset -Z4 a will work with zsh or ksh.
When interpreted as a number, In POSIX sh and printf, if there's a leading zero, it's interpreted as an octal number. So, you need to strip those 0s first:
printf '%04d\n' "${a#"${a%%[!0]*}"}"
Or you could use awk that doesn't have that issue:
awk 'BEGIN{printf "%04d\n", ARGV[1]}' "$a"
