Well explained. but it will be a great help If demo code is provided as I am very new to this. Cant figure out the right syntax to write a code for wait(wrmutex) and signal(wrmutex) and the other signals.
Great video and explanation. It looks like writers might wait forever on readers if the readers keep coming in. If it's true, how would we solve that problem?
That's the problem of starvation. In that case solution can have provision of priority i.e if a process waiting long for it's turn it's priority will be higher and at that time writing process will get it's chance because of it's high priority. There are number of variations for solution, the one discussed in this tutorial is simple one. Welcome to the channel.
You are correct. In order to deal with such kind of situation solution with priority can be taken into account. A reader or writer waited long will get higher priority. So there are different version of the solutions.
We can make the duplicate file and can allow access to the writer in the duplicate file . When the condition readers==0 is met, we assign duplicate file to the original file content and routine process can be done without any starvation for both readers and writers .. For implementation, this concept has to be more dynamic , where it have to handle many write operations simultaneously and have to update the last version of the modified file which seems to be challenging..
Very nice tutorial indeed. I have a query, I cant say if that is a valid query or not but please tell me, suppose there are 3 writers. 1) Write 1 - wait - wrmutex = 0 2) Write 2 - wait - wrmutex = -1 3) Write 3 - wait - wrmutex = -2 Now, when writer 1 signal wrmutex then it will become = -1, So how it will become 0 again now? Please help me.
I'm not sure of this answer. I'll just give a try. When a particular writer calls the wait(wrmutex), say writer 2 calls it when wrmutex=0, (store : int temp1=wrmutex+1), then the particular writer will be notified only when the wrmutex value is 1(equal to temp). Likewise when the writer 3 calls it when wrmutex=-1, (store: int temp2=wrmutex+1), the writer 3 will be notified only if the wrmutex=0. The process continues as the same. Hope it helps :)
Question. Let's say there are 5000 readers and 1 writer. The readers just read endlessly. The writer would be waiting forever, yes? Because before one reader is finished, another starts reading. The reading would never end and the writer cannot ever write. Is this correct?
