Read carefully:

item_size_ok(const size_t nkey, const int flags, const int nbytes) {

passes:

    size_t ntotal = item_make_header(nkey + 1, flags, nbytes,
                                     prefix, &nsuffix);

Then conditionally:

    if (settings.use_cas) {
        ntotal += sizeof(uint64_t);
    }

item_make_header is doing:
nsuffix = (uint8_t) snprintf(suffix, 40, " %d %d\r\n", flags, nbytes - 2);

Then:

return sizeof(item) + nkey + *nsuffix + nbytes;

It's convoluted but shirt.

the lengths are:
key +
1 +
data_bytes +
2b appended to data_bytes for an extra "\r\n" +
stringified rep of the flags + data length
+ 4 bytes for spaces and \r\n (these are carriage returns, one byte each)
+ 8b for CAS if enabled

CAS can be turned off via the -C starttime arg. it takes up 8 bytes.

Example:
 * Key : 2c
 * Val : 28b
 * Flg : 0 (1b)

turns into:
 * Key : 3b
 * Val : 30b
 * Hdr : 4b + 3b == 7b
 * Itm : 56b
 => 96b. which is the cap for slab 1 in a default setup.

It's tough to get it exact for small chunks due to the way the header is
added. You should ballpark or tune the -f value to align with your
observed data.

On Sun, 15 Nov 2015, Nicolas Martinez wrote:

> Hi,
> Is CAS always used?
> If yes, we have to always add 56 bytes to the KEY and VALUE ?
> you don't count FLAGS characters??
>
> I've found that  Flags's size (number of characters) impact the storage.
>
> Example:
>  *  Key : 2 characters = 2 bytes
>  *  Value : 28 characters  = 28 bytes
>  *  FLAGS : 1 characters = 1 bytes
> => 31 bytes
>
> seems to take the same storage as
>  *  Key : 1 characters = 1 bytes
>  *  Value : 28 characters  = 28 bytes
>  *  FLAGS : 2 characters = 2 bytes
> => 31 bytes ... wich is the limit to be stored in Slab1
>
> ok for the /r/n ... should take 4 bytes no?
>
> So, if we count 56 bytes for CAS : 56(cas)+31(key+value+flags)+4(/r/n)= 91
>
> Not good... :(
>
> where I'm wrong ??
>
> Le samedi 14 novembre 2015 23:55:12 UTC+1, Dormando a écrit :
>       The mysql docs don't speak for the main tree... that's their own thing.
>
>       the "sizes" binary that comes with the source tree tells you how many
>       bytes an item will use (though I intend to add this output to the 
> 'stats'
>       output somewhere). With CAS this is 56 bytes.
>
>       56 + 2 + 30 == 88. Class 1 by default (in 1.4.24) is 96 bytes, but the
>       item still ends up in class 2.
>
>       Why is this? (unfortunately?) because memcached pre-renders part of the
>       text protocol into the item header:
>
>       *nsuffix = (uint8_t) snprintf(suffix, 40, " %d %d\r\n", flags, nbytes -
>       2);
>       return sizeof(item) + nkey + *nsuffix + nbytes;
>
>       so the flags + length are getting flattened + \r\n added to the end.
>       Together that's just enough to push it over the edge. It'd also be nice 
> to
>       add a highly optimized numerics printf so I could twiddle options to 
> save
>       a few bytes of memory in objects, but don't get your hopes up for that
>       happening soon :)
>
>       On Sat, 14 Nov 2015, Nicolas Martinez wrote:
>
>       > Add: Memcached version : 1.4.4 (RedHat)
>       >
>       > Le samedi 14 novembre 2015 14:49:37 UTC+1, Nicolas Martinez a écrit :
>       >       Hi, few days i'm reading Memcached documentation and blogs... 
> and i don't understand how objects are stored.
>       >
>       > My test
>       >
>       >       3 slabs : 
>       >
>       >  *  96.0 Bytes
>       >  *  120.0 Bytes
>       >  *  240.0 Bytes
>       > Everywhere, it's told :
>       >  *  if data is < 96 Bytes, it will be stored in Slabs1 (96B)
>       >  *  if data > 96B and < 120B, it will be stored in Slabs2 (120B)
>       >  *  if data > 120B, it will be stored in Slabs3 (240B)
>       >  *  etc.
>       > BUT, for example, when i'm creating an 30B object, it's stored in 
> Slab2 (120B), and NOT in Slab1 (96B).
>       >
>       > External sources:
>       >       For example, the default size for the smallest block is 88 
> bytes (40 bytes of value, and the default 48 bytes for the key and flag 
> data). If the size of the first item you store into the cache is less than 40 
> bytes, then a slab with a block size of 88 bytes is created and the value 
> stored.
>       >       => 
> https://dev.mysql.com/doc/mysql-ha-scalability/en/ha-memcached-using-memory.html
>       >
>       >
>       > WRONG
>       >
>       >       A slab class is a collection of pages divided into same sized 
> chunks. Each slab class is referenced to by its chunk size. So we’ll have 
> Slab class 80kb, Slab class 100kb and so on. When an object needs to be 
> stored, its size determines where it gets stored. So if the object is larger 
> than 80kb but less than 100kb, it gets stored into
>       Slab
>       >       class 100kb. 
>       >       => 
> http://returnfoo.com/2012/02/memcached-memory-allocation-and-optimization-2/
>       >
>       >
>       > WRONG
>       >
>       > How i create an object:
>       >
>       >       data=$(pwgen 30 -c 1)
>       >       echo -e 'set 30 0 3600 30\r\n'$data'\r'| nc ${memcached_server} 
> 11211
>       >
>       >
>       > So, when 30B object is creating : 
>       >  *  key name : 30 = 2 bytes
>       >  *  value: 30 characters = 30 bytes
>       >  *  tags : 0 = 1 bytes
>       > => All = 33 bytes
>       > If i add the default 48b as explained on Mysql website : 33 + 48 = 
> 81B ... so < Slab1 (91B)... but always stored in Slab2 (120B)
>       >
>       > So, the size used to store object in the good Slab is not:
>       >  *  object value size
>       >  *  sum of KEY, VALUE and TAGS in bytes
>       > KEY size : 1 character = 1 B
>       > VALUE size : 1 character = 1 B
>       > TAGS size : 1 character = 1 B
>       >
>       > ... as read everywhere
>       >
>       > So, It seems that: (SUM of KEY+VALUE+TAGS )
>       >  *  For slab1 96.0 Bytes, data stored if <= 31 B (SUM of 2+28+1 )
>       >  *  For slab2 120.0 Bytes, data stored if <= 55 B (SUM of 2+52+1 )
>       >  *  For slab3 152.0 Bytes, data stored if <= 87 B (SUM of 2+84+1 )
>       >  *  For slab4 192.0 Bytes, data stored if <= 126 B (SUM of 3+122+1 )
>       >  *  For slab5 240.0 Bytes, data stored if <= 174 B (SUM of 3+170+1 )
>       >  *  etc.
>       >
>       > My configuration :
>       >  *  Chunk Size : 48
>       >  *  Chunk Growth Factore: 1,25
>       >  *  Max Bytes: 64.0 MBytes 
>       >
>       > So, someone could explain me how the data is stored in the right 
> Slabs???
>       > How calculate it??? 
>       >
>       > Thank you
>       >
>       >
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