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This threshold is currently eight, so if count is less than or equal to 8, the vectored I/O operation occurs in a very memory-efficient manner off of the process’ kernel stack.

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My high school math book uses the word "vectorized" often and never uses vectored.But I guess it is fine for whatever you use because I don't know it indepth anyway. :D – StackUnderblow Aug 4 '11 at 2:04
up vote 4 down vote accepted

I'm going to be presumptuous here and assume that you are not aware of the use of this in the programming sense. (If you are, I don't mean to be condescending.)

A stack in any programming sense is a FILO (First In, Last Out) structure. Think of this as a stack of plates at the buffet table: you can't take a plate in the middle, you must take the plate on top.

Stacks only have 3 operations:

  1. Push - Put something on the top of the stack
  2. Pop - Take something off of the top of the stack
  3. Peek - Look at (but don't touch) something in the middle of the stack

My description of operation #2 has "off of" in it; that's what the original "off of"in your example means. To paraphrase: the I/O operations [pops] operations off in a memory-efficient manner.

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I checked the context, which says that the segments (when small enough) are allocated on the Linux kernel stack, rather than on the specific process' kernel stack. This seems to indicate that small operations are allocated and executed not in the process' kernel stack, but in the Linux kernel stack. So off of in this case means out of, or not in.

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The usage in this instance should be "off" not "off of". The phrase "off of the stack" should only be used for things that are coming from the stack. – Peter Shor Mar 24 '12 at 15:15

"off of" means "from" or "on". It's actually a quite literal phrase.

The sentence is rather confusing. It could be be written as:

"If the count is less than or equal to 8, the vectored I/O operation occurs on the kernel stack of the process, which is very memory-efficient."

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I could be wrong, but I think you're reading the meaning wrongly. The operation wouldn't be any more efficient than usual if it were executed on the process' kernel. The reason why it is efficient is because it is on the Linux kernel. – Daniel Aug 4 '11 at 2:58

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