> Note also that this is the intersection of bits that are available on both ARM and x86. If you want it to be portable, you need both architectures. Just because ARM64 doesn’t use a bit doesn’t mean that x86 doesn’t and vice versa.

Your mask/tag doesn't need to use the same bits on x86 and ARM to be portable, though.

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jandrewrogers 2 days ago | parent [-]

It depends on the application, those bits may be materialized across architectures. The objective was maximizing safety in all contexts.

My perspective is biased by the requirements of high-assurance systems.

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rhdjebejdbd 2 days ago | parent | next [-]

It doesn't depend on the application unless the application shares the same pointers between x86 and arm which doesn't make any sense to me.

Otherwise they're right, it's not the intersection that matters but just the total bits available

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jandrewrogers 2 days ago | parent [-]

Eh? The values you can store in the tags are absolutely dependent on the number of available bits. That’s a simple type safety problem. This requirement is architecture independent.

You can’t cram 8 bits of tag in 7 bits if the latter is all the architecture has available. Hence why you have to design for the smallest reliable target.

	▲	injidup 2 days ago \| parent [-]
		You are agreeing with each other whilst still arguing

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forrestthewoods 2 days ago | parent | prev [-]

If one platform uses the upper 56 bits and another uses the lower 56 bits that doesn’t mean you have 0 bits available for tagging. It means you have 8 bits and have to go through a conversation when moving from one platform to another. This is perhaps annoying but perfectly fine.

Kinda weird to materialize pointers across architectures rather than indices.

But in any case surely the relevant consideration is “fewest number of free pointer bits on any single platform”. And not “intersection of free bits across all platforms”. Right?

	▲	loeg a day ago \| parent [-]
		Right.