semaphores makeWater（）同步 [英] Semaphores makeWater() synchronization

问题描述

这个程序声称解决makeWater（）同步问题。但是，我不明白如何。我是新的信号量。

This program claims to solve makeWater() synchronization problem. However, I could not understand how. I am new in semaphores. I would appriciate if you can help me to understand this code.

推荐答案

因此，你需要使H2O（2Hs和一个O）组合中同时运行的H线程和O线程数。

So you need to make H2O (2Hs and one O) combinations out of number of simultaneously running H-threads and O-threads.

事情是一个O需要两个H。两个不同的水分子之间没有共享。

The thing is one 'O' needs two 'H' s. And no sharings between two different water molecules.

所以假设O和H线程的数量启动他们的进程。

So assume number of O and H threads start their processes.

• 没有O线程可以超越 P（o_wait）锁定，并应等待。


• 一个随机幸运的H线程（说H * -1）可以通过 P（mutex） = 1），并将进入 if（count％2 == 1），然后向上计数'mutex'（现在mutex = 1） $ c> P（h_wait）。 （这个计数实际上是指H计数）


• 因为'mutex'被增加计数，另一个随机H线程（H * -2）将开始通过 P（mutex）（现在mutex = 0和count = 2）。但现在计数是偶数 - >因此它进入 else 。然后它会 V（o_wait）（现在o_wait = 1）并停留在 P（h_wait）。 >
  
• 现在H * -1仍在中的上一个位置，如果阻止。但是因为o_wait被递增计数为1，一个幸运的O线程（O *）可以继续其进程。它将做两个 V（h_wait） s（现在o_wait = 0，h_wait = 2），以便2个以前的H线程可以继续（没有任何其他，现在h_wait = 0 ）。因此，所有3（2个H和O）可以完成其过程，而H * -2增加计数'互斥'（现在mutex = 1）。


• 全局变量完成一个分子后，mutex = 1，h_wait = 0和o_wait = 0，所以正好是初始状态。现在前面的过程将会一次又一次地发生，因此会产生H2O分子。

• No O thread can go beyond P(o_wait) because o-wait is locked, and should wait.
• One random lucky H thread(say H*-1) can go pass P(mutex)(now mutex = 0 and count = 1) and and will go inside if(count%2 == 1), then up-count 'mutex' (now mutex = 1) and block in P(h_wait). (This count actually refers to H count)
• Because 'mutex' was up-counted another random H-thread(H*-2) will start to go pass P(mutex) (Now mutex = 0 and count =2). But now the count is even -> hence it goes inside else. Then it will V(o_wait)(now o_wait = 1) and stuck in P(h_wait).
• Now H*-1 is still at the previous position inside if block. But because o_wait is up-counted to 1, a lucky O thread(O*) can continue its process. It will do two V(h_wait) s (Now o_wait = 0, h_wait = 2), so that 2 previous H threads can continue(No any other, now h_wait = 0). So all 3 (2 Hs and O) can finish its process, while H*-2 is up-counting the 'mutex' (now mutex = 1).
• Now the final values of global variables after completion one molecule, mutex = 1, h_wait = 0 and o_wait = 0, so exactly the initial status. Now the previous process will happen again and again, hence H2O molecules will be created.

我想你会明白。如有任何问题，请提出问题。 :)）

I think you get clear with it. Please raise questions if any. :))

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