In article <2024Sep23.184821@mips.complang.tuwien.ac.at>,   
   Anton Ertl  wrote:   
   >albert@spenarnc.xs4all.nl writes:   
   >>Now look   
   >>   
   >>OK   
   >>MARK-TIME HERE 1,000,000,000 ERASE ELAPSED .mS   
   >>244.344mS  OK   
   >>MARK-TIME HERE 8,000,000,000 ERASE ELAPSED .mS   
   >>67782.522mS  OK   
   >>   
   >>Apparently zeroing a gigabyte takes approximately a quarter of a second.   
   >>   
   >>8 Gigabyte would take 2 seconds. Although the whole program fits   
   >>comfortably in RAM, it takes 30 times longer.   
   >>What gives?   
   >   
   >You obviously have other stuff in more than 8GB of RAM, and paging   
   >enough of that out to accomodate your Forth process takes its time (or   
   >it could also swap out parts of the memory of the Forth system).   
   >That's why I consider swapping pretty useless these days.  If RAM is   
   >too small for your needs, buy more RAM.   
      
   I have 8 Gbyte ram to spare in this situation (total 16).   
      
   However this must be the correct explanation. I never thought that the effect   
   was   
   so dramatic, though. Maybe put 16 Gbyte more solves it.   
      
   Erasing memory is not comparable to a database. The part of the Forth that must   
   be in memory at any given time is kbytes not Mbytes. Linux is apparently not   
   smart   
   about it.   
      
   >                                                      A   
   >- anton   
      
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