8 MB compressed (& if so: zvmlinux.initrd or ramdisk.image.gz?) , or 8 MB uncompressed RAM disk?
If the latter: I've successfully used a RAM disk of 12 or 16 MB. You need to (a) go into your kernel config and set CONFIG_BLK_DEV_RAM_SIZE, and (b) modify whatever it is that creates your RAM disk image (dd + mke2fs perhaps). If the former: you might check the RAM address of your firmware, the RAM runtime address of the boot wrapper (-Ttext ???? in arch/ppc/boot/simple/Makefile), and the size of your RAM. The firmware needs to jump into zvmlinux.initrd, which will copy itself to the -Ttext address, then decompress the kernel to 0 and copy the RAM disk to the end of RAM. You can have problems if any of these things overlap. -----Original Message----- From: Kent Borg [mailto:[EMAIL PROTECTED] Sent: Tuesday, July 01, 2003 1:30 PM To: linuxppc-embedded at lists.linuxppc.org Subject: Large initrd and arch/ppc/boot/simple Question We are booting our kernel via the arch/ppc/boot/simple mechanism to package up our kernel and initrd. It seems to work, until we get too big. What is big? Roughly 8 MB for our one big uncompressed userland program, plus the kernel, bash, busybox, and various userland utilities. Can initrd's get that large and still work? Is the simple boot code sensible with these sizes? (I notice that the "avail ram" message that gets printed is just hard coded number...) Thanks, -kb, the Kent who is running roughly 2.4.21-pre6. ** Sent via the linuxppc-embedded mail list. See http://lists.linuxppc.org/
