Hello On Wed, May 20, 2020 at 08:26:32AM +0200, Uwe Kleine-König wrote: > On Tue, May 19, 2020 at 11:55:55PM -0400, David Dgien wrote: > > When CONFIG_PASSWORD_DEFAULT is unset, the default_passwd buffer is set > > I assume you mean "If CONFIG_PASSWORD_DEFAULT is set to an empty > string". Yes. > > > to the empty string. The read_default_passwd() function wants to read at > > least two characters from that buffer, causing GCC to generate an array > > bounds warning. > > I cannot reproduce that warning. Which gcc version do you use and on > which platform? Mentioning the exact warning in the commit log helps > finding the resulting commit when searching for a fix. arm-none-eabi-gcc --version prints "arm-none-eabi-gcc (Arch Repository) 10.1.0" I found the issue when building for rpi_defconfig and vexpress_defconfig. The warning I get when building from master (commit c10b20dc83ac): barebox/common/password.c: In function 'login': barebox/common/password.c:173:5: warning: array subscript [1, 2147483647] is outside array bounds of 'const char[1]' [-Warray-bounds] 173 | c = buf[i]; | ~~^~~~~~~~ In file included from barebox/common/password.c:30: include/generated/passwd.h:1:19: note: while referencing 'default_passwd' 1 | static const char default_passwd[] = ""; | ^~~~~~~~~~~~~~ I guess the compiler doesn't know that strlen(default_passwd) = 0, just that length > 0 so the most it can assume is that the loop has to consume at least two chars, and the empty string only contains one. > > > Make the default_passwd buffer have at least 2 bytes so > > this warning is not generated. > > > > Since the read_default_passwd() function is only called when > > default_passwd is not the empty string, this is not a functional change. > > I don't understand the problem for the empty password. With > default_passwd = "" we have strlen(default_passwd) = 0 so the for loop > doesn't run at all. Yes, that's correct, which is one reason why this is not functionally different. But the compiler doesn't seem to be smart enough to know that. > > As I understand the code (at commit c10b20dc83ac) for uneven lengths of > default_passwd the last accessed byte is the trailing '\0' and for even > length it's the byte before the trailing '\0'. This should be ok?! > > Am I missing something? When working on this reply, I realized there was another solution I missed when I was trying to find ways to short-circut the compiler previously. If I add: if (ARRAY_SIZE(default_passwd) == 1) return -ENOSYS; in the read_default_passwd() function, that would short-circut the compiler preventing the warning message, and is less hacky. I can resubmit with that change instead. > > Best regards > Uwe > > -- > Pengutronix e.K. | Uwe Kleine-König | > Industrial Linux Solutions | https://www.pengutronix.de/ | Thanks, David Dgien _______________________________________________ barebox mailing list barebox@xxxxxxxxxxxxxxxxxxx http://lists.infradead.org/mailman/listinfo/barebox