On 04/23/2010 03:59 PM, Alexander Graf wrote:
Ah so the 31st bit is optional as far as userspace is concerned? What does it mean? (just curious)
The 0x80000000 bit declares that a pointer is in 24-bit mode, so that applications can use the spare upper bits for random data.
See http://en.wikipedia.org/wiki/31-bit for an explanation.
Interesting. Luckily AMD made the top 16 bits of pointers reserved in
x86-64.
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Do not meddle in the internals of kernels, for they are subtle and quick to panic.
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