On Mon, 16 Aug 2004 22:21:19 +0200, Wolfgang Denk wd@denx.de wrote:
In message cfqlq9$1d9$1@sea.gmane.org you wrote:
I have u-boot working on a custom board based on the MV64360 with a PPC 750GX processor. I've set the autoboot timeout to 3 secs, with the bootcmd to 'go 0x40004'. I have a 'serial_getc' in main_loop so that I can transfer a binary to the 40004 mem. location using a PCI interface via the host machine.
Ummm.. what _exactly_ are you doing? What has a "serial_getc in main_loop" to do with a PCI transfer to memory? and why do you have to do such modifications at all when all can be done using envrionment variables?
Okay, I have a PCI driver on the host machine that can dump stuff into the SDRAM on the board. The serial_getc() in main_loop allows me to put a binary in SDRAM at location 0x40004, 'cos around this time, the SDRAM would've been initialized by u-boot.
An alternate would be to dump the binary on to flash and then within main_loop, copy the binary from flash to RAM and then execute it. (Problem is, I don't have Ethernet devices on the board.) Also, I just have 2 MB of flash, so uImage + initrd + ... will take up more than 2 MB. So I have to do something like this.
On timeout, u-boot tries to transfer control to the binary, but it seems there's some problem:
What exactly is the contents of the memory at 0x00040004 at this time? [and how can you be sure about this?]
I dump the binary via PCI into the SDRAM on the board. On the host side, I specify an offset of 0x40004 and put the binary into /dev/mvsdram0. This is the SDRAM0 BAR. I can read the contents back and verify, they are coherent.
However, jumping to this location (40004) from u-boot via function pointers results in the said trap. If I initialize the function pointer to u-boot code in the RAM, (let's say, main_loop), it works alright. ie, I keep going into main_loop over and over again. So the binary I copied into RAM... either isn't copied (improbable) or the processor doesn't read its contents.
Also, can this be due to some caching? It should not, 'cos there shouldn't have been any access to this location earlier for the data at that location to be cached, but one possibility that I can think of right now.
If there was no access to this location earlier, then how do you think this location could contain executable code?
Yes, because I transferred it through the PCI.
I hope things are a little clearer now..
Best regards,
Wolfgang Denk
Amit.