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[bash]printf octal instead of decimal

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# 1  
Old 04-14-2009
[bash]printf octal instead of decimal
Hello everybody,

I would like to understand why the printf function is returning me an octal value with this command :

printf %4.4d 0010 returns 0008

printf %4.4d 10 returns 0010

Thanks for help.
# 2  
Old 04-14-2009
Because of the '0' at the beginning. This usually tells the system to interpret this as an octal value, much like 0x starts a hex number.
# 3  
Old 04-14-2009
Allright, but removing the zeros is not the solution i m looking for.
Is there a way to tell printf to interpret it as decimal value ?
# 4  
Old 04-14-2009
The obvious question, then, is: why not?
Code:
$ var1="0011"
$ echo $var1
0011
$ echo $var1 | sed -e 's/^0\+//'
11
$ printf "%04d\n" $( echo $var1 | sed -e 's/^0\+//' )
0011
$ var2=$( printf "%04d\n" $( echo $var1 | sed -e 's/^0\+//' ) )
$ echo $var2
0011
$ var2=$( printf "%05d\n" $( echo $var1 | sed -e 's/^0\+//' ) )
$ echo $var2
00011

Pretty simple, isn't it? Other than that (or using bc instead of sed) I know no way to "force" printf on how to interpret the argument.
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