Question about Address Range Validation in Crash Kernel Allocation

[Baoquan He]

Hi,

On 03/18/24 at 12:00pm, chenhaixiang (A) wrote:
> Dear kexec Community Members,
> 
> I encountered an issue while using kexec-tools on my x86_64 machine.
> When there is a segment marked as 'reserved' within the memory range allocated for the crash kernel in /proc/iomem,the output appears as follows:
> 2d4fd058-60efefff : System RAM
>   2d4fd058-58ffffff : System RAM
>     49000000-58ffffff : Crash kernel
>       53cbd000-53ccffff : Reserved

What kernel are you using? the version of kernel, and kexec-tools?

If you are testing on the latest mainline kernel, you could meet the
issue Dave have met and fixed in below patch:

[PATCH] x86/kexec: do not update E820 kexec table for setup_data
https://lore.kernel.org/all/ZeZ2Kos-OOZNSrmO@darkstar.users.ipa.redhat.com/T/#u

Thanks
Baoquan

> 
> The crash_memory_range array will encounter incorrect address ranges:
> CRASH MEMORY RANGES
> 000000002d4fd058-0000000048ffffff (0)
> 0000000053cbd000-0000000048ffffff (1)
> 0000000059000000-0000000053ccffff (0)
> 
> Read the code, I noticed that the get_crash_memory_ranges() function invokes exclude_region() to handle the splitting of memory regions, but it seems unable to properly handle the scenario described above.
> The code logic is as follows:
> ...
> 	if (start < mend && end > mstart) {
> 		if (start != mstart && end != mend) {
> 			/* Split memory region */
> 			crash_memory_range[i].end = start - 1;
> 			temp_region.start = end + 1;
> 			temp_region.end = mend;
> 			temp_region.type = RANGE_RAM;
> 			tidx = i+1;
> 		} else if (start != mstart)
> 			crash_memory_range[i].end = start - 1;
> 		else
> 			crash_memory_range[i].start = end + 1;
> 	}
> ...
> If start < mstart < mend < end, resulting in crash_memory_range[i].end becoming less than crash_memory_range[i].start, leading to incorrect address ranges.
> I would like to know if this behavior is reasonable and whether it is necessary to validate the address ranges for compliance at the end.
> 
> Thank you for your time and assistance.
> 
> Chen Haixiang
> 
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