我试图修改著名的“Dekker's algorithm”,以便您可以同时使用三个进程这是我的代码:
package DekkersAlgorithm;
class DekkerAlg {
/* Iterations done by each Thread */
static final int iterations = 2000000;
/* Shared variable */
static volatile int sharedInteger = 0;
/* P Thread for critical section */
static volatile boolean wantp = false;
/* Q Thread for critical section */
static volatile boolean wantq = false;
/* Z Thread for critical section */
static volatile boolean wantz = false;
/* Thread turn */
static volatile int turn = 1;
class P extends Thread {
public void run() {
for (int i=0; i<iterations; ++i) {
/* No critical section */
wantp = true;
while (wantq || wantz) {
if (turn == 2) {
wantp = false;
while (turn == 2)
Thread.yield();
wantp = true;
}
}
/* Critical section */
++sharedInteger;
/* End critical section */
turn = 2;
wantp = false;
}
}
}
class Q extends Thread {
public void run() {
for (int i=0; i<iterations; ++i) {
/* No critical section */
wantq = true;
while (wantp || wantz) {
if (turn == 1) {
wantq = false;
while (turn == 1)
Thread.yield();
wantq = true;
}
}
/* Critical section */
--sharedInteger;
/* End critical section */
turn = 1;
wantq = false;
}
}
}
class Z extends Thread {
public void run() {
for (int i=0; i<iterations; ++i) {
/* No critical section */
wantz = true;
while (wantp || wantq) {
if (turn == 3) {
wantz = false;
while (turn == 3)
Thread.yield();
wantz = true;
}
}
/* Critical section */
++sharedInteger;
/* End critical section */
turn = 3;
wantz = false;
}
}
}
DekkerAlg() {
Thread p = new P();
Thread q = new Q();
Thread z = new Z();
p.start();
q.start();
z.start();
try {
p.join();
q.join();
z.join();
System.out.println("The value of the sharedInteger is " + sharedInteger);
System.out.println("It should be different from 0.");
}
catch (InterruptedException e) {}
}
public static void main(String[] args) {
new DekkerAlg();
}
}
它在低迭代次数下运行良好,但是当我将这个变量设置为500(+)时,程序有时无法完成。我认为a
livelock
发生在最后两个Threads
之间,但我需要一个关于如何解决它的线索。你能帮帮我吗?
最佳答案
我认为你没有适当地延长。在dekker中,它意味着如果两个人都想进入他们的cs,那么谁就可以进入;这里它似乎意味着如果其他人想进入他们的cs,那么谁就必须等待。对于2个过程来说,它们是直接对立的;对于3个过程来说,就不那么多了。
一种方法是有一个进程列表,指定如果对CS有争用,谁必须等待谁这样,如果P - Q想要进入,Z刚刚退出,Z将被移到列表的结尾,所以你有一个方法在P - Q之间选择(如果你能用一个方法来表示这个“列表”的话,它的修改可能是原子的,这是可行的,因为只有6种不同的模式来表示,那就更好了!)