模擬哲學家就餐

同時提供可能會帶來死鎖的解法和不可能死鎖的解法

可能帶來死鎖

import threading

import time
import random
chopstick=
for i in
range(5
):1)
)class
philosopher
(threading.thread)
:def
__init__
(self,id)
: self.id=
id threading.thread.__init__(self)
defrun
(self)
:while1:
#思考print
('{}號哲學家開始思考'
.format
(self.id)
) time.sleep(random.uniform(
0.1,
0.5)
)#休息
print
('{}號哲學家開始休息'
.format
(self.id)
) time.sleep(random.uniform(
0.1,
0.5)
)#吃飯 i號(0到4)哲學家取i(0到4)號和i-1號筷子
chopstick[self.id]
.acquire(
) time.sleep(
0.3)
#增加死鎖的概率
chopstick[
(self.id-
1)%5
].acquire(
)print
('{}號哲學家開始吃飯'
.format
(self.id)
) time.sleep(random.uniform(
0.1,
0.5)
) chopstick[
(self.id-
1)%5
].release(
) chopstick[self.id]
.release(
)if __name__ ==
"__main__"
:for i in
range(5
):phi=philosopher(i)
phi.start(
)

絕對不可能產生死鎖的解法

限制最多只有4個哲學家拿筷子，則不會產生死鎖

import threading

import time
import random
chopstick=
count=
0#統計拿著筷子的人數，最多只能有4個哲學家拿筷子
for i in
range(5
):1)
)class
philosopher
(threading.thread)
:def
__init__
(self,id)
: self.id=
id threading.thread.__init__(self)
defrun
(self)
:global chopstick,count
while1:
#思考print
('{}號哲學家開始思考'
.format
(self.id)
) time.sleep(random.uniform(
0.1,
0.5)
)#休息
print
('{}號哲學家開始休息'
.format
(self.id)
) time.sleep(random.uniform(
0.1,
0.5)
)if count<4:
#吃飯 i號(0到4)哲學家取i(0到4)號和i-1號筷子
count+=
1 chopstick[self.id]
.acquire(
) time.sleep(
0.5)
chopstick[
(self.id-
1)%5
].acquire(
)print
('{}號哲學家開始吃飯'
.format
(self.id)
) time.sleep(random.uniform(
0.1,
0.5)
) chopstick[
(self.id-
1)%5
].release(
) chopstick[self.id]
.release(
)if __name__ ==
"__main__"
:for i in
range(5
):phi=philosopher(i)
phi.start(
)

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