15 Mar
2007
15 Mar
'07
12:28 a.m.
Hi Roger!
You are right - only very limited hits on google.
Yes, quite right. The 11 gets as operand the address
where to store the program counter for later returning.
Since the unit is 12bit and has 8192 words of memory
it uses two consecutive words for storing the return
address or loading the destination. And a bottleneck
again si, that the return address has to be stored
in the first 128 words of core. I think this is a
severe drawback of this machine: 4 bits for identifying
the instruction and only 7 for the operand. So the
0 and 11 are restricted to the "zero-page"...
Yes. From today's point of view this seems
strange. But considering the sign/value notation
it makes sense somehow.
More information later...
thanks,
Erik.
> Then 12
bit models were the 902, 102C, ARCH 105 and Minim or 12/12.
I doubt you will find much apart from the names in timelines. normal use. Every extra Kg in the Avionics adds a lot
to the overall
takeoff weight. For a helicopter, I think the ratio was somewhere
around six. Then there's the power, every extra watt needs bigger
alternators/generators and a bigger engine to drive them, and more
fuel to carry, which means more engine power and so on.
Never thought about this in
that way. But sounds
absolutely logic.
In 920 assembler, instructions did not have mnemonics,
you used their
decimal operation code values. To indicate 'B Line modification' in
the assembler you wrote a / before the operation code. The assembler
simply added 16 to the opcode, so /8 = 24, but you don't have the B-Line
bit on your machine - its always on.
OK, I understand this now. What looked a typical list
file and of did you specify the operands those days?
Lot of assembly was done by hand, right?
The Programmer
Electronic Control (Let's call it PEC in the future)
stores PC+1 (as your 18bit) to the specified location into
core. The new address is taken from the position in memory
where the index register points to. So in my case this is
a perfect table jump.
Ah, so you do this ?
0 word with address of routine
11 link word Right. But
here again, the bit 11 is the sign bit and
thus the shift does not affect the bit 11 of the
destination. I.e. bit 0 of A is moved to bit 10 of Q
in shifting right for example.
I see. The Q register was always a bit weird.