The Bounded Buffer Problem

Operating System: The Bounded Buffer Problem
Topics discussed:
Classic Problems of Synchronization:
1. The Bounded Buffer Problem.
2. Solution to the Bounded Buffer Problem using Semaphores.
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[@harekrishna4584]

our teacher recommends ur channel as she herself learns here lmaooo

[@Lyricverse116]

😂😂😂

[@yashaswininag3783]

Because it is recommended in VTU syllabus.. You people are unbelievable.. No gratitude towards anyone

[@ChandanKSwain]

Explanation level is outstanding. It makes too easy to grasp...

[@venkateshpatil7109]

Thanks

[@aquadap219]

i watched the past 6 videos of process synchronization and theyre amazing videos thank you!

[@ganesh_hulagur]

Bruh your videos are really help full Plzz do upload remaining topics like *The reader -writer problem *The Dining philosophers problems #monitors etc🙏

[@adelgado0723]

I think this is a version of the reader-writer problem

[@maanyamuthamma3577]

You deserve an award for doing this 🎉🎉

[@mamathaav4627]

Your teaching method is really good.keep on going sir.😊

[@rupambiswas953]

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[@ghaziabadw3494]

thank you so so much for saving me at this last moment

[@jameysiddiqui6910]

This is basic discussion which is good, please can you explain the scenario where producer and consumer can work in paralell.

[@kayakMike1000]

I don't think you really need semaphores and mutexes for this. The producer and consumer could be operating in parallel through some threading mechanism. Assuming there's only one consumer and one producer, we could have a read pointer "chase" the "writer" pointer. the consumer thread isn't allowed to change anything other than it's own read pointer, and we also assume that if the previous element in the buffer has already been consumed. If the consumer measures the read and write pointers to be identical, then the consumer doesn't use that value untill the write pointer is different. If the read and write pointers are different, the consumer consumes the data, then increments its read pointer to point at the next element in the buffer. Now the producer has a write pointer that will increment if the next element in the buffer is not where the read pointer is pointing to. The idea here is the read pointer always chases the write pointer, you could even do this with a ring buffer... With this scheme, the buffer is "empty" when everything has been read, and the read pointer and write pointer point at the same thing. The buffer is full if the write pointer can't advance because the next pointer is where the read pointer actually is..

[@j4jeev678]

Just make 2 thread functions for each producer and consumer and then run both the threads and join it with main thread

[@habibaadawi22]

u are a life saver . thank u so much sir

[@zuhal6900]

That's what i've been looking for!!!! just perfect!!

[@TheEkimjeet]

great presentation and slides! love the content

[@user-ytuser]

thanks you explained an important topic in a easy manner

[@baraaabuasal5626]

i am an electrical engineer i dont know anything about cs but i saw the video, i clicked, i am now studying this problem with you and i understand everything. thank you but may i ask, what is this programming language?

[@pranavvenkata475]

the language used for the implementation of the solution is C

[@putrescentvermin]

This was basically my OS project.

[@Lyricverse116]

Great videos 💯

[@ith3r7]

Well explained !!

[@liorlivyatan]

Why performing signal(mutex) before signal(full)? If a context switch happens in between, won't it cause problems?

[@kayakMike1000]

Why are they using mutexes and semaphores in the first place?

[@-shultz]

yeah , empty should be called before and then the lock must be released .

[@DrJALAGHARI]

A very good way of explanation. Keep it up.

[@advithgowda620]

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[@snehasharma3041]

Amazing explanation 🥹

[@tasnimsart3430]

Thanks so much

[@adelgado0723]

Thank you for the excellent content!

[@darshil-oh2le]

Awesome explaination ✌✌

[@PythonisLove]

great content as always

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😄

[@hmanmusic3908]

your presentations are great keep it up!!

[@arvinhipolito157]

One of the best explanation. Thank you very much

[@kayakMike1000]

I don't think you really need mutex or semaphores if you only have one producer for one consumer...

[@mrymzrzr4741]

Thank you was really helpful

[@kenzakhelkhal7931]

Thank u so much it was very helpful appreciate ur efforts

[@pranavmuthige427]

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[@aryansharma9933]

Thank you!

[@ayahamad5931]

Thanks sir❤❤

[@FarmanOfficial777]

this is called teaching....

[@marcustam4387]

I wonder if semaphores can solve this problem

[@sanathkumarshenoy9955]

Nice one

[@bellaajaziz7204]

can you please upload the remaining topics like deadlocks :)

[@rasikannanl3476]

Fantastic

[@farahmohamed2104]

precise

[@Jamie-fj8dn]

Thxxx sirr

[@hombdr.pathakkshetri8893]

wawwwaaaaa

[@kswethakumari412]

I haven't listened anything before

[@johndoe3092]

Too many semaphores ! Can we do with just 1 ?

[@michaelcenzoprano8173]

i agree, you should only need a mutex lock. You can just maintain a count of elements in the buffer.

[@brahamaggarwal1800]

The thing is when you are dealing with multiple processes that are sharing data then simple if-else statements fail to synchronize our processes that's why it is necessary to use these many semaphores here.

[@modemharikumar9232]

1st comment good lecture

[@satioOeinas]

i dont think this is the correct implementation

[@sricharan8575]

how semaphores solve bounded waiting since u said test.and.set.lock cant solve bounded waiting..coz new process may enter critica.sec while old one was preempted so called the unlucky process?????

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