From: wn0612@d81.de.invalid   
      
   Hi Per,   
      
   sorry for the nitpicking, but ...   
      
   MagerValp schrieb:   
   >>>>>> "S" == Sx/40  <1@2.3> writes:   
   >   
   > S> I had a look around for the very RTFM 38911 question.   
   > S> The thing that I don't find _clear_ is that I found around the wording:   
   >   
   > S> "38911 bytes available for _BASIC_"   
   >   
   > S> does this mean that writing in assembler or in some weird mode lets the   
   > S> programmer user more than 38911 bytes?   
   >   
   > Yes, you can use all 65536 bytes of ram (plus the 1024 nybbles of   
   > color ram).   
      
   ... on the C64 the first two RAM locations $00 and $01   
   are the so named 6510 processor port. The underlaying   
   RAM is not accessible without evil tricks (using sprites,   
   the Commodore REU or open address space accesses).   
   So there are nominally 65534 bytes of RAM available to   
   a machine language programmer. If the programmer needs   
   to use the processor's stack, another couple of bytes   
   from the $0100...$01ff stack area may not be available   
   as ordinary RAM.   
   On the other side a really evil programmer may use some   
   CIA or VIC hardware registers as some sort of RAM, if   
   he doesn't make use of the functions behind. That way   
   you may get a bit more than 65536 bytes available. And   
   don't let us forget the processor registers ;-)   
      
   > Check mapping the c64 and the programmer's reference guide   
   > for information about address 1, which controls memory banking.   
      
   Ups, sorry, it seems I overshoot the mark ;-)   
      
      
   Womo   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)