On 4/8/21 1:06 PM, Frank Rowand wrote:
>>> +#define FDT_ALIGN_SIZE 8
>>> +
>>
>> Use existing define ? Or was that local in libfdt ?
>
> I don't see a define in libfdt. If anyone finds one,
> I'll switch to it.
>
Turns out that was hardcoded in scripts/dtc/libfdt/fdt.c
+ /* The device
On 4/8/21 1:27 PM, Rob Herring wrote:
> On Thu, Apr 8, 2021 at 10:17 AM wrote:
>>
>> From: Frank Rowand
>>
>> The Devicetree standard specifies an 8 byte alignment of the FDT.
>> Code in libfdt expects this alignment for an FDT image in memory.
>> kmemdup() returns 4 byte alignment on openrisc.
Hi Guenter,
Thanks for the review!
On 4/8/21 10:32 AM, Guenter Roeck wrote:
> On 4/8/21 8:17 AM, frowand.l...@gmail.com wrote:
>> From: Frank Rowand
>>
>> The Devicetree standard specifies an 8 byte alignment of the FDT.
>> Code in libfdt expects this alignment for an FDT image in memory.
>>
On Thu, Apr 8, 2021 at 10:17 AM wrote:
>
> From: Frank Rowand
>
> The Devicetree standard specifies an 8 byte alignment of the FDT.
> Code in libfdt expects this alignment for an FDT image in memory.
> kmemdup() returns 4 byte alignment on openrisc. Replace kmemdup()
> with kmalloc(), align
On 4/8/21 8:17 AM, frowand.l...@gmail.com wrote:
> From: Frank Rowand
>
> The Devicetree standard specifies an 8 byte alignment of the FDT.
> Code in libfdt expects this alignment for an FDT image in memory.
> kmemdup() returns 4 byte alignment on openrisc. Replace kmemdup()
> with kmalloc(),
From: Frank Rowand
The Devicetree standard specifies an 8 byte alignment of the FDT.
Code in libfdt expects this alignment for an FDT image in memory.
kmemdup() returns 4 byte alignment on openrisc. Replace kmemdup()
with kmalloc(), align pointer, memcpy() to get proper alignment.
The 4 byte
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