> The kexec system call seems to require that the vmlinuz loading address is
> aligned to a page boundary. 4096 bytes is a fairly common page size, but
> perhaps not the only possibility? Does kexec require additional alignments?
Sorry for the brief description. The problem, more specifically, is
reported by kexec-tools/kexec/kexec.c:343:
/* Verify base is pagesize aligned.
* Finding a way to cope with this problem
* is important but for now error so at least
* we are not surprised by the code doing the wrong
* thing.
*/
if (base & (pagesize -1)) {
die("Base address: 0x%lx is not page aligned\n", base);
}
https://git.kernel.org/pub/scm/utils/kernel/kexec/kexec-tools.git/tree/kexec/kexec.c?id=HEAD#n343
A more pressing issue at the moment though is that initramfs in vmlinuz is
ignored when booting. Is it supposed to work with a MIPS kernel?
Fredrik
> --- a/arch/mips/boot/compressed/calc_vmlinuz_load_addr.c
> +++ b/arch/mips/boot/compressed/calc_vmlinuz_load_addr.c
> @@ -44,12 +44,8 @@ int main(int argc, char *argv[])
> vmlinux_size = (uint64_t)sb.st_size;
> vmlinuz_load_addr = vmlinux_load_addr + vmlinux_size;
>
> - /*
> - * Align with 16 bytes: "greater than that used for any standard data
> - * types by a MIPS compiler." -- See MIPS Run Linux (Second Edition).
> - */
> -
> - vmlinuz_load_addr += (16 - vmlinux_size % 16);
> + /* The kexec system call requires page alignment. */
> + vmlinuz_load_addr += (4096 - vmlinux_size % 4096);
>
> printf("0x%llx\n", vmlinuz_load_addr);
>
