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Full Discussion: [bash]printf octal instead of decimal
Top Forums Shell Programming and Scripting [bash]printf octal instead of decimal Post 302306886 by pludi on Tuesday 14th of April 2009 06:38:44 AM
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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LEARN ABOUT OSX

echo

ECHO(1) 						    BSD General Commands Manual 						   ECHO(1)

NAME

     echo -- write arguments to the standard output

SYNOPSIS

     echo [-n] [string ...]

DESCRIPTION

     The echo utility writes any specified operands, separated by single blank (' ') characters and followed by a newline ('
') character, to the
     standard output.

     The following option is available:

     -n    Do not print the trailing newline character.  This may also be achieved by appending 'c' to the end of the string, as is done by iBCS2
	   compatible systems.	Note that this option as well as the effect of 'c' are implementation-defined in IEEE Std 1003.1-2001
	   (``POSIX.1'') as amended by Cor. 1-2002.  Applications aiming for maximum portability are strongly encouraged to use printf(1) to sup-
	   press the newline character.

     Some shells may provide a builtin echo command which is similar or identical to this utility.  Most notably, the builtin echo in sh(1) does
     not accept the -n option.	Consult the builtin(1) manual page.

EXIT STATUS

     The echo utility exits 0 on success, and >0 if an error occurs.

SEE ALSO

     builtin(1), csh(1), printf(1), sh(1)

STANDARDS

     The echo utility conforms to IEEE Std 1003.1-2001 (``POSIX.1'') as amended by Cor. 1-2002.

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