31 --- 28* you already used 2x bits, so it's 31--28 not 27. still tho on my MIPS sheet, for some reason, the jumpAddress is assigned with PC+4[31-28}, address, 2'b0 i don't know if it's a typo or what, but theyre using the PC + 4 rather than the PC.
Basically, to address 4GB of memory in J encoding, you need to address 2^32 bits, represented by 32 bits. In MIPS, which is a fixed legnth microarchitecture, we only have 32 bits per instruction. 6 bits goes to the opcode, remaining 26 bits goes to the addresses in J encoding. The question is, how do you address something that needs 32 bits minimum with only 26 bits for addressing? We have to make up for 6 bits one way or another. Recovering 2 of these bits from the lower order bits is easy, as a word is minimum 4 bits. To get the remaining 4 bits out of the missing 6, we take it from the program counter. Now we have a way to represent 32 bit addressing with only 26 bits!
Thanks for this video sir, found it very very helpful...
Glad it helped!
Thank you, great explanation
Glad you found it helpful and thanks for the comment and support.
Great explanation, thanks a lot!
31 --- 28* you already used 2x bits, so it's 31--28 not 27.
still tho on my MIPS sheet, for some reason, the jumpAddress is assigned with PC+4[31-28}, address, 2'b0
i don't know if it's a typo or what, but theyre using the PC + 4 rather than the PC.
and it's [31 -- 28] [27 --- 2] [1 0]
from 2 ---> 27 there is 26 digits.
you got the idea, dnt be goofy
Life Saver
Thank you ❤❤
You're welcome 😊
what is the the program counter instruction number ? i thought it meant the opCode but now im not so sure
Ig opCode would come before these 26b's
Basically, to address 4GB of memory in J encoding, you need to address 2^32 bits, represented by 32 bits.
In MIPS, which is a fixed legnth microarchitecture, we only have 32 bits per instruction. 6 bits goes to the opcode, remaining 26 bits goes to the addresses in J encoding.
The question is, how do you address something that needs 32 bits minimum with only 26 bits for addressing? We have to make up for 6 bits one way or another. Recovering 2 of these bits from the lower order bits is easy, as a word is minimum 4 bits.
To get the remaining 4 bits out of the missing 6, we take it from the program counter.
Now we have a way to represent 32 bit addressing with only 26 bits!
Not sure about your video. Haven't watched you before. Is this assembler code, or another language?
MIPS assembly which is a RISC architecture.
@@ProfessorHankStalica thank you.
@@ProfessorHankStalica Simply amazing !!!Hands down best one I've found