Hello Studying the ARM/Versatile port of U-Boot in order to understand booting and initialization nuances.
As per documentation from http://infocenter.arm.com/ SDRAM on this platform is selected by CS#0 and gets mapped in 0x0000_0000 - 0x0800_0000. Versatile configuration defines TEXT_BASE=0x0100_0000, so it is 16Mb away from 0x0; Linux port of Versatile defines loading address of the kernel as 0x0000_8000, resulting in 32kB gap form the beginning of RAM. I understand that 16Mb gap is to let kernel image to fit in RAM, but isn't it too large, seems a waste of space to me? Then why do we need to keep 32KB space, what's the point? As U-Boot image usually occupies ~200KB, wouldn't it be easier to allocate ~8Mb room, which would be quite enough for the U-Boot, its environment, malloc area, stack etc., Linux kKernel then starts and reclaims the memory of U-Boot. Would appreciate if someone explains me this nuance. Thanks in advance! -- Roman Mashak _______________________________________________ U-Boot mailing list U-Boot@lists.denx.de http://lists.denx.de/mailman/listinfo/u-boot
