Extremely helpful. Seems very simple and has not changed for so many years but still i get confused time to time so review of the video was great refresher.
Can you explain how you arrived at that statement? This figure has 30 gates. A standard ripple carry would take 30 stages. The figure in this video has 10+3 (=13) stages for the max delay.
Assuming your question is "why do we start with way 0 (and not 1)" the answer is pretty simple, there is no reason why we couldn't have start with way 1 and then went with way 0. I would say though that we went with what is generally consider "conventnion."
Off the top of my head, I'm not sure what you are refering to by "CNF method." If you can expand a little, I likely would be able to respond more helpfully.
At first you say that Neither branch is taken but at the end the last one you tell us that again Neither of the branch is taken. I am confused can you help me understand if i get something wrong?
i don't understand how I get from 0x070 to the tag of 0000001 and set 1 offset 100000 etc. i.ex.i don't get the translation from the cache access address to the cache address
My first thought is that you are dealing with two different "nomenuclatures", one being hexadecimal and the other being binary. Have you been keeping that it mind?
I have a question. What if the addresses in memory are not all uniform bit size. Each of your addresses in the example are 12 bit. But what if they differ, such as between 12 - 16 bits? Will the process you are using here work on addresses of varying bits? Such as 0x408ed4, 0x10019d94, 408ed8...
Assuning you are refering to immediate values (bit positation 15-0 for the immediate values portion) they are all 16 bits. It may only show a smaller number if the higher order bits are all zero. As a further extendion, in MIPS, all addresses are 32 bits (with immediate values of 16 bits). Does that address your question?
I don't know why in your example with the same tag we always have the same set ? What will be happen when we have the addresses with the same tag and the different set ?
can anyone please explain why we initially set the valid bit to 1 or 0. Every video I come across "so we set this bit to 1" with no further reasoning behind it. Otherwise everything was super helpful!
Extremely helpful. Seems very simple and has not changed for so many years but still i get confused time to time so review of the video was great refresher.
So helpful!!!
Thank you so much!!
thank you tony stark
If we dont have the last line, what the pipeline will be? Can we begin the IF of the first loop line directly in circle 14?
The last line, as I interpret it anyway, is never executed, so removing it really wouldn't change anything.
@@matthewwatkins88 I see. Thank you very much!
this seems longer in terms of prop delay than just using ripple carry?
Can you explain how you arrived at that statement? This figure has 30 gates. A standard ripple carry would take 30 stages. The figure in this video has 10+3 (=13) stages for the max delay.
This is so cool
Thanks for helping out!
Isn't it 13 instructions in the loop? (including 5 to 17)....or did he include the 18th cycle?
It does say it takes 13 (17) cycles to complete the loop (so I'm unsure about the issue)
@@matthewwatkins88 yup just realized...Great work!!
Excellent video. Than you so much.
GREAT VIDEO THANK YOU helped me with my lab 3 project :)
why do we go with way 0 first?
Assuming your question is "why do we start with way 0 (and not 1)" the answer is pretty simple, there is no reason why we couldn't have start with way 1 and then went with way 0. I would say though that we went with what is generally consider "conventnion."
Would this grouping method work for the CNF method?
Off the top of my head, I'm not sure what you are refering to by "CNF method." If you can expand a little, I likely would be able to respond more helpfully.
Still helping me in 2024 - big thanks!
true
Thank you so much! This tutorial was a lifesaver.
(Y)
one of the best explanations
You guys rock! This video answers ‘The abstract’ 😍
GOD BLESS YOU FOR THIS!!!💪
nice
I think in the third case you meant first branch (beq $t0,$0, end) is taken only
I'll stop my head, I would agree with you.
Great video, much better explained than my professor.
He didn’t even explain he just told u what to do 😤
At first you say that Neither branch is taken but at the end the last one you tell us that again Neither of the branch is taken. I am confused can you help me understand if i get something wrong?
This video shows all three cases. So you will learn about all three options by watching this video.
@@matthewwatkins88 When the exercise says that the branch is followed i should think that the branch is taken or not taken?
At one instant in time the branch is either taken or not taken. But the video shows both options so you will learn about both cases from the video.
@@matthewwatkins88 Thanks a lot.
Hey, thank you very much for you great example, I was trying to remember logic circuits, and that was so helpful
thank you for saving my life
i don't understand how I get from 0x070 to the tag of 0000001 and set 1 offset 100000 etc. i.ex.i don't get the translation from the cache access address to the cache address
My first thought is that you are dealing with two different "nomenuclatures", one being hexadecimal and the other being binary. Have you been keeping that it mind?
❤😊
Very good content!!!! Thank you very much!
ur a life saver
Please speak in hindi.
I have a question. What if the addresses in memory are not all uniform bit size. Each of your addresses in the example are 12 bit. But what if they differ, such as between 12 - 16 bits? Will the process you are using here work on addresses of varying bits? Such as 0x408ed4, 0x10019d94, 408ed8...
Assuning you are refering to immediate values (bit positation 15-0 for the immediate values portion) they are all 16 bits. It may only show a smaller number if the higher order bits are all zero. As a further extendion, in MIPS, all addresses are 32 bits (with immediate values of 16 bits). Does that address your question?
@@matthewwatkins88 yes ok thank you very much for your answer!
What happens when blocking and nonblocking assignments are intermixed?
In what way would mixing blocking and non-blocking assignments cause issues?
👍👍👍👍👍
Thank you for both videos. Better than what I got at the university.
Teşekkürler verdiğiniz bilgi için.
goat no one explains it better and got a final tomorrow wish me luck
Absolutely amazing and helpful!
great example, thank you!
5:18 Doesnt the LRU change to 1 because of Hit for last address ??
Save me from the poor lecture notes but advanced question sheet
Where can i buy this??
You would do much better than my current cse230 teacher. thank you.
thanks a lot man
the table of addr,tag,set,offset of the left is memory or cache? The table on the right side is cache right?
What is the purpose of the offset bits? I’m confused on that point.
The offset bits are the least significant portion of the number. The number as a whole has three portions, the offset, the set, and the tag.
I don't know why in your example with the same tag we always have the same set ? What will be happen when we have the addresses with the same tag and the different set ?
Oh now i understood we must look at the set first before we compare the tags
thank you for your effort
Useful video, thanks!
can anyone please explain why we initially set the valid bit to 1 or 0. Every video I come across "so we set this bit to 1" with no further reasoning behind it. Otherwise everything was super helpful!
By the nature of the system every bit must have a value, which in our case must be either a zero or a one. Does that help any?
Thank you Sir!