Bonsoir Paul,

> There is a bug in the web page: there is only 3 bytes reserved at offset 5

Sorry I have not paid attention to that, but as you can see
in my mail, I have written that you can ignore 3 bytes :-)

> Also you seems to have closed your eyes at a bug in FreeDOS:
> 
> yes, 1c0 is the size, but not in bytes, but in paragraphs

Yes. The size is given in paragraphs, all DOS versions do
that because otherwise MCB could not allocate > 64 kB. Maybe
I should have explained that instead of just mentioning the
page which describes the data structure.

> I am still at that point at reading your message.

As mentioned, manipulating the memory allocation chain by
hand is quite a bit of work, but for the sake of experiment,
it is acceptable. Also remember that because your system will
always have the same state after booting with the same options,
you can write down the commants and later put them into a text
file to re-apply the trick as often as you want, for example
to check with different apps whether they still crash. As you
cannot do interactive things, tha would be:

a 39b:3
dw 5463

a 57ff:0
db 'M'
dw 8
dw 47ff
db 0
dw 0
db 'RESERVE'
db 0

q

But of course the values 39b and 5463 will differ on your system!

Maybe the 47ff also differs, if you have less than 640 kB of low
DOS RAM: Sometimes a few kB are reserved for BIOS purposes at the
end, like at 9fc0:0, so there would be a MCB structure at 9fbf:0
instead of 9fff:0 which my example assumes.

Either way, you would write everything from "a 39b..." to "q" into
a text file, then you can do this:

debug < textfile.txt

to apply all changes with a single command. To find the correct
values instead of 39b, 5463 and 47ff, you would use the manual
MCB chain inspection and calculations outlined in my mail. The
starting point is the value shown by MEM /D /P as MEM's location.

It is a technical thing, but you will understand after playing
around with the concept for a bit :-)

Regards, Eric



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