From: rick.c.hodgin@gmail.com   
      
   On Wednesday, October 17, 2018 at 8:31:32 PM UTC-4, sdn4...@nosp   
   cedham.gmail.com wrote:   
   > For RH:   
   > Dos allocates all of RAM for a COM program by default.   
   > When a DOS program exits properly, then RAM is the same for the next   
   > program upon entry minus whatever a memory manager does to move around   
   > DOS (and we need not care)   
   > and drivers/tsrs. All memory upon entry for the new program is the   
   > same and contiguous from an ASM viewpoint unless the previous program exited   
   > incorrectly and left a memory leak.   
      
   There's no way to know how much memory is there though.  I don't   
   ever recall seeing a function to return the size of a memory block,   
   but only functions that would fail if you tried to allocate too   
   much, returning a value in the BX register indicating how many para-   
   graphs were actually available to allocate, which would tell you how   
   much was left.   
      
   That always seemed like a fundamental flaw in DOS's memory allocation   
   scheme.  But, knowing what I know now about how these things work, I   
   may not have understood it correctly back then.  I can't see such a   
   glaring failure existing.  I must not have had enough knowledge of   
   how to obtain the actual allocated size.   
      
   --   
   Rick C. Hodgin   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)