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sandbox: Add un/map_sysmen() to deal with sandbox's ram_buf

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Sandbox doesn't actually provide U-Boot access to the machine's physical
memory. Instead it provides a RAM buffer of configurable size, and all
memory accesses are within that buffer. Sandbox memory starts at 0 and
is CONFIG_DRAM_SIZE bytes in size. Allowing access outside this buffer
might produce unpredictable results in the event of an error, and would
expose the host machine's memory architecture to the sandbox U-Boot.

Most U-Boot functions assume that they can just access memory at given
address. For sandbox this is not true.

Add a map_sysmem() call which converts a U-Boot address to a system
address. In most cases this is a NOP, but for sandbox it returns a
pointer to that memory inside the RAM buffer.

To get a U-Boot feature to work correctly within sandbox, you should call
map_sysmem() to get a pointer to the address, and then use that address for
any U-Boot memory accesses.

Signed-off-by: Simon Glass <sjg@chromium.org>
This commit is contained in:
Simon Glass 2013-02-24 17:33:14 +00:00
parent e101550a9a
commit 4213fc2913
4 changed files with 32 additions and 0 deletions
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12
include/common.h
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@ -870,6 +870,18 @@ int cpu_disable(int nr);
int cpu_release(int nr, int argc, char * const argv[]);
#endif
/* Define a null map_sysmem() if the architecture doesn't use it */
# ifndef CONFIG_ARCH_MAP_SYSMEM
static inline void *map_sysmem(phys_addr_t paddr, unsigned long len)
{
return (void *)(uintptr_t)paddr;
}
static inline void unmap_sysmem(const void *vaddr)
{
}
# endif
#endif /* __ASSEMBLY__ */
#ifdef CONFIG_PPC