Hi, Bash's printf appears to ignore the \c backslash escape:
$ printf "before \c after \a" before \c after $ $ type printf printf is a shell builtin Of course the Open Group's description of printf (http://www.opengroup.org/onlinepubs/000095399/utilities/printf.html) does not specify that \c in the format string should be interpreted, but the builtin echo, and the echo and printf from coreutils, do interpret it: $ echo -e "before \c after \a" before $ $ type echo echo is a shell builtin $ /bin/echo -e "before \c after \a" before $ $ /usr/bin/printf "before \c after \a" before $ When \c is provided via the %b conversion specifier, it is interpreted: $ printf "before %b after \a" '\c' before $ It makes it seem odd that a straight \c isn't. (Of course, a straight \c is less useful than an indirect one, but still.) $ bash --version GNU bash, version 3.2.39(1)-release (i686-pc-linux-gnu) Copyright (C) 2007 Free Software Foundation, Inc. Benno
