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IBM 2030 Manual Of Instruction page 77

Processing unit, field engineering

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Assume: I.
That the Problem Program Tokes 5,000 Bytes and Will Begin at
Location 2048.
2.
That the Supervisor is in Locations 0000 - 2047 and Has a
Storage Key of 15 and a Protection Key of 0.
, - - - - - - - ,
Problem
Set Storage
Key of 2048-
4095 to I
Set Storage
Key of 4095-
6143 to I
Set Storage
Key of 6143-
8191 to I
t
_ _ _ _ Program is
Read Into
Loc. 2048-7074
A Storage Key of I Was Chosen for This Problem
Program. Actually Any Key from 1-14 Could
Have Been Used. Zero Would Offer No Protec-
tion, While 15 Is Already Being Used By the
Supervisor Program.
"Assembled"
PSWWould
Probably
be
Like This
I. System
Mask
of All Ones to Allow Interrupts.
2. Protection Key of I to Match the Storage Key
--
....
-
Associated with This Program.
Control
3.AMWPfield
~
°
I
°
I~
Allow /
Jnning
Prob.
-
-
-+- Passes to the
Prob lem Prog.
Machine
State
State
Interrupt
4. Instruction Address of 2048.
Problem {
Program
6144 - 8191
III
4096 - 6143
IT1
2043 - 4095
iii
Supervisor Program
~
0000 - 2047
Figure
1-31.
USing Storage Protection
The protection key in the PSW used by
the supervisor program is zero.
This
allows the supervisor program to modify
data anywhere in main storage.
The main
storage area occupied by
t~e
supervisor
program has a storage key of
15.
This
means that unless a problem program has
a key in its PSW of
0
or
15,
it will not
be
able to modify or change information
in the area being used by the supervisor
program.
This is unlikely because it is
the supervisor program that assigns
storage and protection keys to the prob-
lem program.
Each block of
2048
bytes does not
have to have a different number set in
its storage key.
However, each program
in main storage should have a different
storage key aSSigned to protect one
program from another.
For instance, the
supervisor program may take up one block
of 2048 bytes which is aSSigned a stor-
age key of
15.
This storage key would
most likely
be
assigned by the supervi-
sor program just after it is read into
the system.
The problem program is then
read into the machine (as a result of a
section of the supervisor program) •
This program takes up 3 blocks of 2048
bytes.
Each of these three blocks is
assigned the same storage key
(1,
for
example) by the supervisor program.
The
PSW for the problem program is given a
protection key that matches its storage
keys.
This allows the problem program
to alter itself if necessary, but pre-
vents it from altering another program.
MULTIPROGRAMMING
So far, we have only discussed the con-
cept of two programs in the computer:
a superv1sor program and problem
program.
There are two or more problem
1-73

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