[vox-tech] lame question on memory allocation

[Ryan Zeeturt]

A word is 2 bytes I think on x86. Then after that there is double-word,
quad-word and paragraph, and more nerdy names I'm sure.

Word-aligned means that you can't have a memory address starting at
anything that isn't a multiple of 2. So for different types of alignments,
let's say it were aligned to N bytes, then you couldn't have a memory
address
starting at anything that isn't MEM_ADDRESS mod N.


-ryan


On Tue, 21 Jan 2003, Peter Jay Salzman wrote:

>
> you'd think by now i'd know stuff like this.  i'm embarrased to have to
> ask this, but here it goes.
>
> i'm reading the man page for electric fence, and i'm not fully
> understanding the sections on EF_ALIGNMENT and "WORD-ALIGNMENT AND
> OVERRUN DETECTION".   i feel like i "almost" understand them.
>
> i think i understand the concept of memory page as being the minimum
> chunk of memory the kernel handles internally (8192 bytes minimum
> allocation of memory on x86) and alignment, but i guess i don't know
> what a word is.
>
> for example, the man page says that malloc() may be required to return
> word aligned memory pages, so in the diagram:
>
>
>          1 page allocated by malloc()
> x       ------------
> x+1     |          |
> x+2     | 8192     |
>         | bytes on |
>         |   x86    |
>         |          |
>         ------------
>
> i guess that places a restriction on what "x" is, but because i don't
> know what a word is, i don't know what that restriction is.
>
> what's a word?  :)
>
> or does it mean that there's a restriction on *size* of the page and not
> the starting point?
>
> pete
>
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> then you win. -- Gandhi, being prophetic about Linux.
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