tag 14299 notabug thanks On 04/28/2013 12:44 PM, Pavel Elkind wrote: > Dear developers, > > I found the following potential bug in printf (version 8.17). > > Actual result: > `printf "\\n"` prints a newline caracter.
Of course. That's what POSIX requires it to print. $ set -x $ printf ".\\n." + printf '.\n.' . . $ set - > > Expected result: > `printf "\\n"` prints a sequence of two individual characters, '\' and 'n', > like '\n', but not a newline character. If you want printf to print a literal backslash, you have to properly escape it. There are two levels of escaping to consider; shell escaping (before printf ever sees its argv), and printf escaping. You missed a level, because you forgot that within "", the shell converts \\ into a literal \ as part of the argv, and as my 'set -x' trace showed above, you were passing only one backslash, not two, to printf. Within printf, when it sees the single backslash-n sequence, it converts that escape sequence to newline. You probably meant to do any one of these equivalent actions: printf '.\\n.' printf .\\\\n. printf ".\\\\n." all of which result in the argv handed to printf still containing two backslashes. As such, I'm closing this as not a bug, although you may continue to reply here if you have further comments. -- Eric Blake eblake redhat com +1-919-301-3266 Libvirt virtualization library http://libvirt.org
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