Subject: lib/22019: printf C99 conformance with %#.o
To: None <firstname.lastname@example.org>
From: Johan Danielsson <email@example.com>
Date: 06/30/2003 15:49:33
>Synopsis: printf C99 conformance with %#.o
>Arrival-Date: Mon Jun 30 13:50:02 UTC 2003
>Originator: Johan Danielsson
>Release: NetBSD 1.6T
<The following information is extracted from your kernel. Please>
<append output of "ldd", "ident" where relevant (multiple lines).>
System: NetBSD blubb.pdc.kth.se 1.6T NetBSD 1.6T (BLUBB) #531: Thu May 29 13:09:25 CEST 2003 firstname.lastname@example.org:/usr/misc/src/netbsd/netbsd-cvs/src/sys/arch/i386/compile/BLUBB i386
Currently printf("%#.o", 0) produces the empty string (""), and this
is correct in terms of C89, but C99 includes this piece of text: "if
the value and precision are both 0, a single 0 is printed". This
indicates that the output should be "0".
But I'm not sure I think this makes much sense. For instance "%#.x"
would still result in "".
So either we should fix printf to comply with C99, or we should decide
that this is just rubbish. I have not seen a rationale for C99, so I
don't know why this change was made.
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