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Modern Operating Systems by Herbert Bos and Andrew S. Tanenb...
Modern_Operating_Systems_by_Herbert_Bos_and_Andrew_S._Tanenbaum_4th_Ed.pdf
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Modern Operating Systems by Herbert Bos and Andrew...
Modern_Operating_Systems_by_Herbert_Bos_and_Andrew_S._Tanenbaum_4th_Ed.pdf-M ODERN O PERATING S YSTEMS
Modern Operating Systems by Herbert...
Modern_Operating_Systems_by_Herbert_Bos_and_Andrew_S._Tanenbaum_4th_Ed.pdf-M ODERN O PERATING S YSTEMS
Page 155
124
PROCESSES AND THREADS
CHAP. 2
while (TRUE) {
while (TRUE) {
while (turn != 0)
/
*
loop
*
/;
while (turn != 1)
/
*
loop
*
/;
critical
region( );
critical
region( );
turn = 1;
turn = 0;
noncritical
region( );
noncritical
region( );
}
}
(a)
(b)
Figure 2-23.
A proposed solution to the critical-region problem. (a) Process 0.
(b) Process 1. In both cases, be sure to note the semicolons terminating the
while
statements.
finds it to be 0 and therefore sits in a tight loop continually testing
turn
to see when
it becomes 1.
Continuously testing a variable until some value appears is called
busy waiting
.
It should usually be avoided, since it wastes CPU time. Only when
there is a reasonable expectation that the wait will be short is busy waiting used.
A
lock that uses busy waiting is called a
spin lock
.
When process 0 leaves the critical region, it sets
turn
to 1, to allow process 1 to
enter its critical region. Suppose that process 1 finishes its critical region quickly,
so that both processes are in their noncritical regions, with
turn
set to 0.
Now
process 0 executes its whole loop quickly, exiting its critical region and setting
turn
to 1.
At this point
turn
is 1 and both processes are executing in their noncritical re-
gions.
Suddenly, process 0 finishes its noncritical region and goes back to the top of
its loop. Unfortunately, it is not permitted to enter its critical region now, because
turn
is 1 and process 1 is busy with its noncritical region. It hangs in its
while
loop
until process 1 sets
turn
to 0.
Put differently, taking turns is not a good idea when
one of the processes is much slower than the other.
This situation violates condition 3 set out above: process 0 is being blocked by
a process not in its critical region. Going back to the spooler directory discussed
above, if we now associate the critical region with reading and writing the spooler
directory, process 0 would not be allowed to print another file because process 1
was doing something else.
In fact, this solution requires that the two processes strictly alternate in enter-
ing their critical regions, for example, in spooling files. Neither one would be per-
mitted to spool two in a row.
While this algorithm does avoid all races, it is not
really a serious candidate as a solution because it violates condition 3.
Peterson’s Solution
By combining the idea of taking turns with the idea of lock variables and warn-
ing variables, a Dutch mathematician, T. Dekker, was the first one to devise a soft-
ware solution to the mutual exclusion problem that does not require strict alterna-
tion. For a discussion of Dekker’s algorithm, see Dijkstra (1965).
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