Improve the documentation of the memory block size growth feature

[nical]

Followups from #254. cc @MarijnS95

[MarijnS95]

Thanks, that's quick! I think there's one comment that you can still address: #254 (comment). For the others, I marked most as resolved and the few remaining ones require larger refactors that protrude the scope of the original PR.

[MarijnS95]

Just noting that for min_device_memblock_size you say "GPU only memory type", here you say "shared". That probably doesn't mean "shared with CPU" but "shared with multiple smaller allocations", right?

[nical]

Yes, I meant shared as opposed to dedicated blocks which hold a single allocation and are not limited to max_*_memory_block_size. Let me know if I should call it something else.

[nical]

I'm going to make the wording more consistent between the min and max docs, remove "shared" there and drop a note about these limits applying only to shared-as-in-non-dedicated blocks in the struct-level doc.

[nical]

Apologies for the many typos and awkward wording. It looks like in the end I didn't save you much time by submitting the PR myself.

[MarijnS95]

No worries, I would have made the same typos and weird wording myself, you just get blind to it when constantly shifting code around. It always helps to have a second pair of eyes :)

[MarijnS95]

Thanks!

[MarijnS95]

Not something to fix in this PR, nor to mention in the documentation, but I think there's currently a mistake where we allocate such a "dedicated memory block" directly also if the minimum size didn't yet grow to the maximum size:

gpu-allocator/src/vulkan/mod.rs

Lines 472 to 478 in ba9a268

	let memblock_size = allocation_sizes.get_memblock_size(is_host, self.active_general_blocks);
	let size = desc.requirements.size;
	let alignment = desc.requirements.alignment;
	let dedicated_allocation = desc.allocation_scheme != AllocationScheme::GpuAllocatorManaged;
	let requires_personal_block = size > memblock_size;

But fixing that in whichever way (i.e. let requires_personal_block = size > allocation_sizes.max_xxx_memblock_size;) might only make the code more confusing, as we'd have to either immediately allocate all blocks up until a block that would fit, or find the smallest multiple between min..max where that allocation would fit (that's a shared and not dedicated block).