t*g
2 楼
如果W2上Box 12里面没有内容的话,是不是退税时不能拿HSA作为deductible了?
我自己给HSA寄过支票,算作2013 contribution。这部分能不能作为deductible呢?
我这样情况,退税时还需要准备8889吗?
谢谢!
我自己给HSA寄过支票,算作2013 contribution。这部分能不能作为deductible呢?
我这样情况,退税时还需要准备8889吗?
谢谢!
t*8
3 楼
祖传的。
d*w
6 楼
The amount on W2 Box 12 is for employer contribution (including cafeteria
plan) only bah.
Your own contribution would not show there. That's why you need to file 8889
to get deduction.
【在 t******g 的大作中提到】
: 如果W2上Box 12里面没有内容的话,是不是退税时不能拿HSA作为deductible了?
: 我自己给HSA寄过支票,算作2013 contribution。这部分能不能作为deductible呢?
: 我这样情况,退税时还需要准备8889吗?
: 谢谢!
plan) only bah.
Your own contribution would not show there. That's why you need to file 8889
to get deduction.
【在 t******g 的大作中提到】
: 如果W2上Box 12里面没有内容的话,是不是退税时不能拿HSA作为deductible了?
: 我自己给HSA寄过支票,算作2013 contribution。这部分能不能作为deductible呢?
: 我这样情况,退税时还需要准备8889吗?
: 谢谢!
W*o
8 楼
这帖子太长了,粗看了一下。还是通过提问讨论学习一下吧。
有几个问题:
1. 如果几个processes要access shared data structure, 怎么避免race ?
2. 一个 process 对 common data structure 操作结束,怎么通知下一个 process? 按
照什么顺序?
3. 最后一个 process 处理完 common data structure, 怎么告诉其他 processes to
move
on ?
【在 b*******s 的大作中提到】
: http://kukuruku.co/hub/cpp/lock-free-data-structures-the-inside
有几个问题:
1. 如果几个processes要access shared data structure, 怎么避免race ?
2. 一个 process 对 common data structure 操作结束,怎么通知下一个 process? 按
照什么顺序?
3. 最后一个 process 处理完 common data structure, 怎么告诉其他 processes to
move
on ?
【在 b*******s 的大作中提到】
: http://kukuruku.co/hub/cpp/lock-free-data-structures-the-inside
t*d
10 楼
甘肃大头,有一眼。 可以拍个边齿。 BTW,楼下有卖,
http://www.mitbbs.com/article_t/Collectibles/31336969.html
http://www.mitbbs.com/article_t/Collectibles/31336969.html
t*t
11 楼
1. usually lock-free structures use atomic operations and (possibly) retry t
o guarantee race-free conditions without lock. how? read the article. are yo
u waiting others to spoon feed you...?
2. there is no lock so there is no notification. while one thread gain acces
s, the others just keep retrying.
3. since it's the last process, usually the other process already move on. i
guess you are thinking about barrier...
按
to
【在 W***o 的大作中提到】
: 这帖子太长了,粗看了一下。还是通过提问讨论学习一下吧。
: 有几个问题:
: 1. 如果几个processes要access shared data structure, 怎么避免race ?
: 2. 一个 process 对 common data structure 操作结束,怎么通知下一个 process? 按
: 照什么顺序?
: 3. 最后一个 process 处理完 common data structure, 怎么告诉其他 processes to
: move
: on ?
o guarantee race-free conditions without lock. how? read the article. are yo
u waiting others to spoon feed you...?
2. there is no lock so there is no notification. while one thread gain acces
s, the others just keep retrying.
3. since it's the last process, usually the other process already move on. i
guess you are thinking about barrier...
按
to
【在 W***o 的大作中提到】
: 这帖子太长了,粗看了一下。还是通过提问讨论学习一下吧。
: 有几个问题:
: 1. 如果几个processes要access shared data structure, 怎么避免race ?
: 2. 一个 process 对 common data structure 操作结束,怎么通知下一个 process? 按
: 照什么顺序?
: 3. 最后一个 process 处理完 common data structure, 怎么告诉其他 processes to
: move
: on ?
C*1
12 楼
据说在民国年间就有大量制假,祖传的也不一定是真的。
t*t
13 楼
你还不如看看c++ concurrency in action那本书. 写得不错.
【在 b*******s 的大作中提到】
: http://kukuruku.co/hub/cpp/lock-free-data-structures-the-inside
【在 b*******s 的大作中提到】
: http://kukuruku.co/hub/cpp/lock-free-data-structures-the-inside
t*8
14 楼
呵呵,家里传了三代了。不管真假,我这枚都不卖,要不老爹要找麻烦了。有空上个边
齿照片。谢谢大家指教,学了不少知识。
齿照片。谢谢大家指教,学了不少知识。
W*o
15 楼
谢谢,你说的第二点有点像busy waiting/spinning; 如果是一个分布式系统,这样快
有很高的 network communication overhead 吧
今天晚上没心情静下心来读,明天白天有空再详细读一下;
PS,如果这个idea很有创新,原作者应该会投给期刊发表出来了吧?呵呵
t
yo
acces
i
【在 t****t 的大作中提到】
: 1. usually lock-free structures use atomic operations and (possibly) retry t
: o guarantee race-free conditions without lock. how? read the article. are yo
: u waiting others to spoon feed you...?
: 2. there is no lock so there is no notification. while one thread gain acces
: s, the others just keep retrying.
: 3. since it's the last process, usually the other process already move on. i
: guess you are thinking about barrier...
:
: 按
: to
有很高的 network communication overhead 吧
今天晚上没心情静下心来读,明天白天有空再详细读一下;
PS,如果这个idea很有创新,原作者应该会投给期刊发表出来了吧?呵呵
t
yo
acces
i
【在 t****t 的大作中提到】
: 1. usually lock-free structures use atomic operations and (possibly) retry t
: o guarantee race-free conditions without lock. how? read the article. are yo
: u waiting others to spoon feed you...?
: 2. there is no lock so there is no notification. while one thread gain acces
: s, the others just keep retrying.
: 3. since it's the last process, usually the other process already move on. i
: guess you are thinking about barrier...
:
: 按
: to
t*8
16 楼
照边齿是个技术活啊,试了好多次。好像没法贴再回复里。没这个选项啊。
t*t
17 楼
yes, busy waiting is essential for lock-free structures. no one says busy wa
iting should be used everywhere though.
and of course the idea is not new. in fact, every lock is internally impleme
nted with some kind of lock-free structures, using atomic operations. ultima
tely hardware guarantees an operation is atomic.
【在 W***o 的大作中提到】
: 谢谢,你说的第二点有点像busy waiting/spinning; 如果是一个分布式系统,这样快
: 有很高的 network communication overhead 吧
: 今天晚上没心情静下心来读,明天白天有空再详细读一下;
: PS,如果这个idea很有创新,原作者应该会投给期刊发表出来了吧?呵呵
:
: t
: yo
: acces
: i
iting should be used everywhere though.
and of course the idea is not new. in fact, every lock is internally impleme
nted with some kind of lock-free structures, using atomic operations. ultima
tely hardware guarantees an operation is atomic.
【在 W***o 的大作中提到】
: 谢谢,你说的第二点有点像busy waiting/spinning; 如果是一个分布式系统,这样快
: 有很高的 network communication overhead 吧
: 今天晚上没心情静下心来读,明天白天有空再详细读一下;
: PS,如果这个idea很有创新,原作者应该会投给期刊发表出来了吧?呵呵
:
: t
: yo
: acces
: i
t*8
18 楼
Here is the links to the edge photos, sorry no idea how to post photos in
the reply posts.
https://plus.google.com/photos/107125557867417784715/albums/
6002539036801715089/6002539041892124578?pid=6002539041892124578&oid=
107125557867417784715
https://plus.google.com/photos/107125557867417784715/albums/
6002539036801715089/6002539041892124578?pid=6002539041892124578&oid=
107125557867417784715
the reply posts.
https://plus.google.com/photos/107125557867417784715/albums/
6002539036801715089/6002539041892124578?pid=6002539041892124578&oid=
107125557867417784715
https://plus.google.com/photos/107125557867417784715/albums/
6002539036801715089/6002539041892124578?pid=6002539041892124578&oid=
107125557867417784715
g*e
19 楼
test and set指令
w*g
23 楼
这种东西学术界折腾有几年了。Lock在性能上其实没那么不济。Lock free主要还是为
了程序写起来方便,性能上可能还不如lock。
【在 b*******s 的大作中提到】
: http://kukuruku.co/hub/cpp/lock-free-data-structures-the-inside
了程序写起来方便,性能上可能还不如lock。
【在 b*******s 的大作中提到】
: http://kukuruku.co/hub/cpp/lock-free-data-structures-the-inside
t*8
24 楼
就一枚。边齿的照片在上面。请大牛掌掌眼,把把关。四十年前我第一次看见它 的时
候,好像没这么氧化。在想要不要拿碱水洗洗。
候,好像没这么氧化。在想要不要拿碱水洗洗。
y*i
28 楼
一眼真的甘肃大头,硬要说个版别的话,这个是“不断华大嘉禾”,这种瞄一眼就知道
的东西就不用看边齿了。
的东西就不用看边齿了。
t*8
30 楼
@yuanshikai 你太牛了! 长知识了。 膜拜,膜拜中。。。俺是个绝对的新手。多谢赐
教!
教!
t*8
32 楼
自学了一下,这好像是所谓的“开口贝”版。
k*g
48 楼
昏特。
本质上就是根据各任务的时间性,选择适当的保护方案。
Inefficiency = (average time per operation including lock overhead) divided
by (average time per operation excluding lock overhead) minus 1.0
The numerator is the actual performance. The divisor is the time spent "
doing useful stuff".
Ideally inefficiency = 0, i.e. zero overhead. In practice, make inefficiency
small.
Corollary: If the task takes longer time, it's okay to use lock with larger
overhead per-operation, because the inefficiency won't increase much.
nano-scale : lock-free, together with immutable/copyable types.
micro-scale : OS locks.
milli-scale : DB transactions
second-scale : need to show something to the user e.g. sandglass or spinning
beachball
本质上就是根据各任务的时间性,选择适当的保护方案。
Inefficiency = (average time per operation including lock overhead) divided
by (average time per operation excluding lock overhead) minus 1.0
The numerator is the actual performance. The divisor is the time spent "
doing useful stuff".
Ideally inefficiency = 0, i.e. zero overhead. In practice, make inefficiency
small.
Corollary: If the task takes longer time, it's okay to use lock with larger
overhead per-operation, because the inefficiency won't increase much.
nano-scale : lock-free, together with immutable/copyable types.
micro-scale : OS locks.
milli-scale : DB transactions
second-scale : need to show something to the user e.g. sandglass or spinning
beachball
b*s
49 楼
总结得不错
divided
inefficiency
larger
【在 k**********g 的大作中提到】
: 昏特。
: 本质上就是根据各任务的时间性,选择适当的保护方案。
: Inefficiency = (average time per operation including lock overhead) divided
: by (average time per operation excluding lock overhead) minus 1.0
: The numerator is the actual performance. The divisor is the time spent "
: doing useful stuff".
: Ideally inefficiency = 0, i.e. zero overhead. In practice, make inefficiency
: small.
: Corollary: If the task takes longer time, it's okay to use lock with larger
: overhead per-operation, because the inefficiency won't increase much.
divided
inefficiency
larger
【在 k**********g 的大作中提到】
: 昏特。
: 本质上就是根据各任务的时间性,选择适当的保护方案。
: Inefficiency = (average time per operation including lock overhead) divided
: by (average time per operation excluding lock overhead) minus 1.0
: The numerator is the actual performance. The divisor is the time spent "
: doing useful stuff".
: Ideally inefficiency = 0, i.e. zero overhead. In practice, make inefficiency
: small.
: Corollary: If the task takes longer time, it's okay to use lock with larger
: overhead per-operation, because the inefficiency won't increase much.
n*t
50 楼
you apparently used very little MPSC queues in real life, at least in
performance critical settings.
One thing I think you probably do not know is, lock-free algos typically
make things slower, do you know why?
【在 p*u 的大作中提到】
:
: lock-free queues are very important, esp multiple-producer-single-consumer
: and single-producer-single-consumer queues...
performance critical settings.
One thing I think you probably do not know is, lock-free algos typically
make things slower, do you know why?
【在 p*u 的大作中提到】
:
: lock-free queues are very important, esp multiple-producer-single-consumer
: and single-producer-single-consumer queues...
c*4
51 楼
晕死。打酱油地弱弱的问一句,搞清楚茴香豆的茴字有多少种写法真能赚到钱么?有多
少职务是要自己实现同步的。
【在 b*******s 的大作中提到】
: http://kukuruku.co/hub/cpp/lock-free-data-structures-the-inside
少职务是要自己实现同步的。
【在 b*******s 的大作中提到】
: http://kukuruku.co/hub/cpp/lock-free-data-structures-the-inside
p*u
52 楼
w*w
55 楼
1000个线程都不睡 那不得1000个core啊 这个lock free应该是在low contention,
short critical session 时有用吧 并且数据结构得够简单 update 得isolated. CAS
又是个hack 总之是费力不讨好 多年前听过 现在还有人做吗?
short critical session 时有用吧 并且数据结构得够简单 update 得isolated. CAS
又是个hack 总之是费力不讨好 多年前听过 现在还有人做吗?
o*0
59 楼
学习了
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