Re: IMX6UL: Booting kernel with initramfs blocks

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Am Freitag, den 26.10.2018, 16:25 +0200 schrieb Patrick Boettcher:
> On Fri, 26 Oct 2018 13:24:38 +0200
> > Sascha Hauer <s.hauer@xxxxxxxxxxxxxx> wrote:
> > >   Starting kernel at 0x888e1000, oftree at 0x8a542000...
> > 
> > From what I can see these addresses look fine. You have an image
> > that's 28MiB in size, we put it to roughly 5x size into memory and the
> > devicetree behind it. The kernel should then unpack itself to the
> > start of SDRAM. Looks good and It seems barebox actually jumps to the
> > Kernel.
> > 
> > Next thing to try would be enabling CONFIG_DEBUG_LL,
> > CONFIG_DEBUG_UNCOMPRESS and CONFIG_EARLY_PRINTK in the Kernel. Pass
> > "earlyprintk" as kernel option and let's see what happens then.
> 
> (earlyprintk yet to be tried)
> 
> We made a step forward: Reducing the size of the image (from 28 down to
> 17 MiB) makes it boot correctly.
> 
> Not sure where the limitation can come from, if you have an idea.
> 
> Thanks again for the help so far, I will post a follow-up if the
> root-cause is discovered.

My guess is that the kernel is just not able to find the start of RAM
for relocation in that case. With such a fat image and our pessimistic
placement at 5 times that size, the kernel ends up in a region above
the first 128MB of memory.
The kernel needs to be placed in the low 128MB, as this is the address
range that gets masked off from the location in order to find the start
of RAM.

Regards,
Lucas

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