Doubt_in_Addressing [closed]

Question

Though its a trivial doubt but i want to ask Here in this question :

Given is 36bit virtual address means We can address every bit which of address space with 36bits now considering bytes we just need 33bits Right? Because each 8bit will be grouped Here Its mentioned that $2^{^{36}}Bytes$.
So please comment on this.And correct me !

Comments

36 bit virtual address generated by CPU and the set of all logical addresses is logical address space which is 2^36 words.

If byte addressable system then 2^36 bytes where each byte is addressed.

Similarly for set of all physical addresses which is physical address space is 2^30 bytes.

What is your doubt now?

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Given 36bits=64Gb=8GB Thus to address each bytes we need 33bits how are you saying its 2^36 bytes?

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Read about 'byte addressable memory'

https://en.wikipedia.org/wiki/Byte_addressing

https://www.doc.ic.ac.uk/~eedwards/compsys/memory/index.html

https://www.cs.umd.edu/class/sum2003/cmsc311/Notes/CompOrg/simpMem.html

let suppose you have 3 bit of virtual address space , then we know 3 bit will result in 8 combinations and therefor 8 addresses can ge generated by a 3 bit virtual address.now when memory is byte addressable , every combination will correspond to 1B so total addrress space then become 8B in this case.We can suppose memory as array of bytes , when we have 8B of memory then think it like array of 8 elements , in which every entry will correspond to one address , so there are 8 addresses. to represent these 8 addresses , 3 bit virtual address is enough and each address will point to single byte only. in other words 3 bit can address 8 Bytes means 8 addresses.

if memory is 'bit addressable' then each address wil store a bit. So 3 bit means 3 addresses.

https://stackoverflow.com/questions/42903026/what-is-the-main-difference-between-byte-addressable-and-bit-addressable

it was actually a good doubt. if you think i went wrong somewhere tell me.

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Thanks @Rupendra. Actually i was misinterpreting the term byte in the case of virtual memory. Thanks for clearing Got it now.