Thanks for the clear explanation and for sharing ! I have one question, hope you answer: the paging conversion table should be implemented by using high speed hardware, am I right?
@madhura it is 0 -4194303, 4194304 are total numbers that can be represented. 0 is included. same reason a byte, which translates to 256 decimal can only include numbers 0-255. 2^3 is 8 but bit representation 111 is 7.
It has to do with the logical addressing. D effectively tells the CPU how many bits of memory it will require. So since D contains processes 3, 4, 5 and 7 (which are 4 pages... each taking up 4 MB) it tells the CPU "hey I need the memory space 0x0000000h to 0x0D00000h in order for 3,4,5 and 7 to fit in memory". Hope that helps
Hi, you say 11 1111 1111 1111 1111 1111 (22 one's) is 4MB, but it converts into 4194303 in decimal. However, 4MB in decimal is actually 4194304. Can any of you please explain why?
With 22 bits you can represent 4194304 numbers, but the range goes from 0 to 4194303 (zero is included in the range). 1111 converted to decimal is 15, but you can represent 16 numbers with 4 bits (range 0 to 15)! (2 ^ 22) - 1 = 4194303 (2 ^ 4) - 1 = 15
You made this seem as though it was a lot easier than it appears so thanks man! Really helped me out.
I want to smash that like button.Thank you.
Great. Very discreet and coherent. Please share more videos. Thanks.
Thanks for the clear explanation and for sharing ! I have one question, hope you answer: the paging conversion table should be implemented by using high speed hardware, am I right?
Great video as always. Your volume is quite low. Thanks for the videos!
@madhura it is 0 -4194303, 4194304 are total numbers that can be represented. 0 is included. same reason a byte, which translates to 256 decimal can only include numbers 0-255. 2^3 is 8 but bit representation 111 is 7.
9:16 how did we assume that D's perspective is only the addresses from 0x0000000h to 0x0D00000h ? What does this even mean ?
It has to do with the logical addressing. D effectively tells the CPU how many bits of memory it will require. So since D contains processes 3, 4, 5 and 7 (which are 4 pages... each taking up 4 MB) it tells the CPU "hey I need the memory space 0x0000000h to 0x0D00000h in order for 3,4,5 and 7 to fit in memory". Hope that helps
Thank you.Great explanation.
Great explaination！ Thanks
better than my teacher, thank you
Hi, you say 11 1111 1111 1111 1111 1111 (22 one's) is 4MB, but it converts into 4194303 in decimal. However, 4MB in decimal is actually 4194304. Can any of you please explain why?
With 22 bits you can represent 4194304 numbers, but the range goes from 0 to 4194303 (zero is included in the range). 1111 converted to decimal is 15, but you can represent 16 numbers with 4 bits (range 0 to 15)!
(2 ^ 22) - 1 = 4194303
(2 ^ 4) - 1 = 15
Awesome video sir
well explained, thanks
big thanks sir!
godbless
Thank you!