Just checking for certainty. I've run a program that reads a section of ROM 
in my PC. I looked at the file produced by the program I used, and it 
indicated the bytes there as FF for the whole area of ROM that a reading was 
taken of.  In asci, a capital Y with two dots above. Does FF mean that this 
ROM area is unwritten on? The ROM area in question needs be free for a 
(vintage) HD set-up program (or pionter) to reside there. (Prior to the 
change in the video card I've just made, the old card wrote code all over 
the ROM area, thus interfering with the HD setup. Thanks.

"Rich"  wrote in message
news:7lisihF3durp2U1@mid.individual.net...
> Just checking for certainty. I've run a program that reads a section of
> ROM in my PC. I looked at the file produced by the program I used, and it
> indicated the bytes there as FF for the whole area of ROM that a reading
> was taken of.  In asci, a capital Y with two dots above. Does FF mean that
> this ROM area is unwritten on? The ROM area in question needs be free for
> a (vintage) HD set-up program (or pionter) to reside there. (Prior to the
> change in the video card I've just made, the old card wrote code all over
> the ROM area, thus interfering with the HD setup. Thanks.

ROM and RAM. Er, someone gave me a program called GETROM. Without checking
with the guy who gave me it at this point, I'm sure it took a reading of
my RAM memory. Because we wanted to know what was at C800:0000 onwards.
So, I must have taken a snap-shot of my RAM.

The message 
from "Rich"  contains these words:

> Just checking for certainty. I've run a program that reads a section of ROM 
> in my PC. I looked at the file produced by the program I used, and it 
> indicated the bytes there as FF for the whole area of ROM that a
> reading was 
> taken of.  In asci, a capital Y with two dots above. Does FF mean that this 
> ROM area is unwritten on? The ROM area in question needs be free for a 
> (vintage) HD set-up program (or pionter) to reside there. (Prior to the 
> change in the video card I've just made, the old card wrote code all over 
> the ROM area, thus interfering with the HD setup. Thanks. 

 Unprogrammed bit locations in a PROM (then ditto for the later EPROM
technology) would show as being set to '1'. Programming a PROM or EPROM
simply turned those bits that needed to be set to zero into zeros,
leaving those that needed to be set to '1' alone.

 The orginal PROM technology relied on 'blowing' fusable links (hence
the expression 'Blowing or burning a PROM' (burning, as in burning a
fuse out). Obviously a one shot process (apart from still being able to
turn unprogrammed bits into zeroes if that that's all an 'edit'
required).

 EPROMs followed the same rules (it's a TTL thing). The difference, of
course, being that the chip could be 'washed' back to the 'all ones'
state by exposure to an appropriate dosage of UV light via its 'quartz'
window.

 Nowadays, the EEPROM (now referred to as 'flash ram') has ousted the
traditional EPROM chip which, in its turn, has ousted the traditional
PROM chip in the form of a much cheaper all plastic encapsulated
windowless EPROM). It has the charm that it can be programmed in the
more commonly available Eprom programmer but, due to the absence of an
erasure window, like the PROM it substitutes for, cannot be erased to
permit another reprogramming cycle.

 Incidently, finding large blocks of memory set to 1s in the memory map
is often referred to as "Write Only" memory, aka, empty memory slots.
;-)

-- 
Regards, John.

 Please remove the "ohggcyht" before replying.
The address has been munged to reject Spam-bots.

"Johnny B Good"  wrote in message 
news:31303030373730364AF6B4EC36@plugzetnet.co.uk...
> The message 
> from "Rich"  contains these words:
>
>> Just checking for certainty. I've run a program that reads a section of 
>> ROM
>> in my PC. I looked at the file produced by the program I used, and it
>> indicated the bytes there as FF for the whole area of ROM that a
>> reading was
>> taken of.  In asci, a capital Y with two dots above. Does FF mean that 
>> this
>> ROM area is unwritten on? The ROM area in question needs be free for a
>> (vintage) HD set-up program (or pionter) to reside there. (Prior to the
>> change in the video card I've just made, the old card wrote code all over
>> the ROM area, thus interfering with the HD setup. Thanks.
>
> Unprogrammed bit locations in a PROM (then ditto for the later EPROM
> technology) would show as being set to '1'. Programming a PROM or EPROM
> simply turned those bits that needed to be set to zero into zeros,
> leaving those that needed to be set to '1' alone.
>
> The orginal PROM technology relied on 'blowing' fusable links (hence
> the expression 'Blowing or burning a PROM' (burning, as in burning a
> fuse out). Obviously a one shot process (apart from still being able to
> turn unprogrammed bits into zeroes if that that's all an 'edit'
> required).
>
> EPROMs followed the same rules (it's a TTL thing). The difference, of
> course, being that the chip could be 'washed' back to the 'all ones'
> state by exposure to an appropriate dosage of UV light via its 'quartz'
> window.
>
> Nowadays, the EEPROM (now referred to as 'flash ram') has ousted the
> traditional EPROM chip which, in its turn, has ousted the traditional
> PROM chip in the form of a much cheaper all plastic encapsulated
> windowless EPROM). It has the charm that it can be programmed in the
> more commonly available Eprom programmer but, due to the absence of an
> erasure window, like the PROM it substitutes for, cannot be erased to
> permit another reprogramming cycle.
>
> Incidently, finding large blocks of memory set to 1s in the memory map
> is often referred to as "Write Only" memory, aka, empty memory slots.
> ;-)
>
> -- 
> Regards, John.
>
> Please remove the "ohggcyht" before replying.
> The address has been munged to reject Spam-bots.

The program GETROM was, I think, devised specifically to look at my RAM 
memory, to aid in spotting ROM BIOS signatures. That confused me a bit and 
got me talking abour ROM when I should be talking bout RAM. :c) I'm looking 
at byte values in RAM and for some reason I have a whole swath of bytes with 
a hex value of FF.

Anyway, essentially that RAM space I'm looking at is clear, because what's 
in RAM (FF's) has not interfered with a program that needs the space at the 
addresses in question.

Rich wrote:
> 
> The program GETROM was, I think, devised specifically to look at my RAM 
> memory, to aid in spotting ROM BIOS signatures. That confused me a bit 
> and got me talking abour ROM when I should be talking bout RAM. :c)

What it'll actually give you is a mixture of RAM and ROM (or at least 
RAM-shadowed ROM). It's the same address space, and depending on your 
hardware, some of it will be ROM-mapped. If a particular range is full 
of FFs, it's probably RAM-mapped and free.

The standard mapping for the first megabyte is something like this:

00000-9FFFF:	mapped to RAM
A0000-BFFFF:	mapped to video RAM
C0000-DFFFF:	mapped to add-in card BIOSes
E0000-FFFFF:	mapped to motherboard BIOS

Space above C0000 not used by BIOS code is mapped to RAM.


-- 
John Jordan

The message <YuJJm.104570$KU2.96111@newsfe26.ams2>
from John Jordan  contains these words:

> Rich wrote:
> > 
> > The program GETROM was, I think, devised specifically to look at my RAM 
> > memory, to aid in spotting ROM BIOS signatures. That confused me a bit 
> > and got me talking abour ROM when I should be talking bout RAM. :c)

> What it'll actually give you is a mixture of RAM and ROM (or at least 
> RAM-shadowed ROM). It's the same address space, and depending on your 
> hardware, some of it will be ROM-mapped. If a particular range is full 
> of FFs, it's probably RAM-mapped and free.

> The standard mapping for the first megabyte is something like this:

> 00000-9FFFF:	mapped to RAM
> A0000-BFFFF:	mapped to video RAM
> C0000-DFFFF:	mapped to add-in card BIOSes
> E0000-FFFFF:	mapped to motherboard BIOS

> Space above C0000 not used by BIOS code is mapped to RAM.

 Actually, I forgot to mention that unwritten to dram will show as
blocks of data alternating between all ones and all zeros (0x00 and
0xFF). The size of each block depends on the number of address pins
employed by the chips.

 Remember, a 10 address pin chip gives a total of 20 address lines (the
address is strobed in twice, a row address followed by a column
address). In this example (a 1Mbit or 1M nybble, or 1MB depending on the
data bus width of the chip) you would expect to see the block size at
1KB.

 The reason for this effect is on account that half the sense/refresh
amplifiers invert the signal to and from each memory cell to minimise
noise effects. IOW, the charge stored on the gate of the memory cell
either represents a 1 or a 0 depending on its location. At power up, all
the memory cells are in the 'no charged' state, hence that pattern until
real data has been written (whether it's the unprogrammed space of a rom
that's been shadowed into the ram or program data outside of the shadow
memory region).

 HTH

-- 
Regards, John.

 Please remove the "ohggcyht" before replying.
The address has been munged to reject Spam-bots.