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What does the transient and volatile keywords do?

When your not sure consult the 'Bible', 'Java™ Language Specification'
http://java.sun.com/docs/books/jls/t...ses.html#78119


8.3.1.3 transient Fields
Variables may be marked transient to indicate that they are not part of
the persistent state of an object.

If an instance of the class Point:

class Point {
int x, y;
transient float rho, theta;
}

were saved to persistent storage by a system service, then only the
fields x and y would be saved. This specification does not specify
details of such services; see the specification of java.io.Serializable
for an example of such a service.

8.3.1.4 volatile Fields
As described in §17, the Java programming language allows threads to
access shared variables. As a rule, to ensure that shared variables are
consistently and reliably updated, a thread should ensure that it has
exclusive use of such variables by obtaining a lock that,
conventionally, enforces mutual exclusion for those shared variables.

The Java programming language provides a second mechanism, volatile
fields, that is more convenient than locking for some purposes.
A field may be declared volatile, in which case the Java memory model
(§17) ensures that all threads see a consistent value for the variable.

If, in the following example, one thread repeatedly calls the method one
(but no more than Integer.MAX_VALUE times in all), and another thread
repeatedly calls the method two:

class Test {
static int i = 0, j = 0;
static void one() { i++; j++; }
static void two() {
System.out.println("i=" + i + " j=" + j);
}
}

then method two could occasionally print a value for j that is greater
than the value of i, because the example includes no synchronization
and, under the rules explained in §17, the shared values of i and j
might be updated out of order.

One way to prevent this out-or-order behavior would be to declare
methods one and two to be synchronized (§8.4.3.6):

class Test {
static int i = 0, j = 0;
static synchronized void one() { i++; j++; }
static synchronized void two() {
System.out.println("i=" + i + " j=" + j);
}
}

This prevents method one and method two from being executed
concurrently, and furthermore guarantees that the shared values of i and
j are both updated before method one returns. Therefore method two never
observes a value for j greater than that for i; indeed, it always
observes the same value for i and j.

Another approach would be to declare i and j to be volatile:

class Test {
static volatile int i = 0, j = 0;
static void one() { i++; j++; }
static void two() {
System.out.println("i=" + i + " j=" + j);
}
}

This allows method one and method two to be executed concurrently, but
guarantees that accesses to the shared values for i and j occur exactly
as many times, and in exactly the same order, as they appear to occur
during execution of the program text by each thread. Therefore, the
shared value for j is never greater than that for i, because each update
to i must be reflected in the shared value for i before the update to j
occurs. It is possible, however, that any given invocation of method two
might observe a value for j that is much greater than the value observed
for i, because method one might be executed many times between the
moment when method two fetches the value of i and the moment when method
two fetches the value of j.

See §17 for more discussion and examples.

A compile-time error occurs if a final variable is also declared volatile.

--

posted on 2008-03-06 00:35 lqx 阅读(227) 评论(0)  编辑  收藏 所属分类: java


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