Quoting Frank Adam to Adam Fitzpatrick:



FA>Had to rewrite it to put it through Pacific C, but the idea does not

FA>work there, so it seems it's the RTL doing the shuffle not the OS.   

 

AF>That's understandable; if you malloc'ed a 2-byte variable (don't ask

AF>_why_ you would...) DOS would use 32 bytes. Just a bit inefficient.



FA>Wouldn't that be 16 bytes then ? Maybe you meant two 1 byte vars...??



It needs 1 paragraph that you're actually allowed to use and another

paragraph to store the information about who owns that block of memory,

how big it is, and whether or not there is any more memory after it. DOS

keeps a record of what's free and what's been allocated in a linked list

of these Memory Control Blocks. Each one contains a flag to specify whether

or not it's the last one in the chain, the number of paragraphs in this

block, the address of the PSP of the program that owns this block of

memory, and (in at least DOS 5.0+, and possibly 4.0) the name of the program.

 

AF>FWIW, a block of memory allocated by DOS has its size stored at

AF>(s-1):3, where s is the segment returned in AX by the mov ah,48h; int



FA>Yes, that reflects what i've found in Borland, doesn't explain why

FA>Pacific didn't come up with the same results though.



It might just allocate the whole damn chunk of free memory in one go.

It's possible that Borland allocates a new chunk of RAM every time it

needs some, and does its own management of those chunks. That's probably

a good idea if you're in an environment where other programs are competing

for your 640k, but I can't think of many examples of that happening to be

honest. It's still a good idea though.

 

AF>21h call. So memory allocated by DOS will have the (char)(ptr-16) as

AF>'M' or 'Z' and (short)(ptr-13)<<4 will be the size of the memory block

AF>(shifted 4 bits to the left because it's stored as a number of

AF>paragraphs, ie 16-byte blocks).



FA>What determines the size of a paragraph ? Or is it always 16 ?



It's 16 by definition; DOS uses it because the real mode segments are 16

bytes apart (ie 0:0 is 16 bytes below 1:0).



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