CODEWITHSUNDEEP
architecture
user2810581
user2810581

Reputation: 51

Byte / Word Addressable

I am currently studying basic ccomputer architecture and require help solving the following:

How many bits are required to address a 4M x 16 main memory if:

a) The memory is byte addressable?

b) The memory is word addressable?

The answer is, like in most cases located in the back of the text book, but I want to know how to work it out.

Thanks in advance!

Upvotes: 1

Views: 17033

Answers (4)

dildo
dildo

Reputation: 1

That is incorrect Individual bits are not addressable. Admittedly, the memory contains

4M * 16 bits = 2^22 * 2^4 = 2^26 bits

but it is the byte or word count that is used for the addressing.

Upvotes: 0

Iłya Bursov
Iłya Bursov

Reputation: 24145

4M x 16 = 64Mb of memory = 67108864 bytes or 33554432 words (word is usually 2 bytes)

this means that last byte in memory will have address: 0x3FFFFFF (67108864-1) = 26 bits

last 2 byte word will have address: 0x1FFFFFF (67108864/2-1 = 33554432-1) = 25 bits

and if we consider 4 byte word - we have latest address 0xFFFFFF (67108864/4-1) = 24 bits

UPDATE: I wrongly multiplied 4M by 16, actually 16 here is number of bits, so correct calculations will be:

4M x 16 = 4 194 304 cells, each cell is 16 bit = 2 bytes

To address each byte, you need 4 194 304 * 2, so last byte will have address (0x800000-1) = 7FFFFF, ie 23 bits

To address each word (2 bytes), you need 4 194 304, so last word will have address 0x3FFFFF, ie 22 bits

Upvotes: -1

Zhanwen Chen
Zhanwen Chen

Reputation: 1463

If you are working on Chapter 4, No. 5 of Null and Lobur, here's the answer:

a) 23 bits

4M * 16 = 2^2 * 2^20 * (2^4 / 2^3) (16 bits / 8 bits is a byte) = 2^2 * 2^20 * 2^1 = 2^23 => 23 bits.

b) 22 bits

Assuming a word is 16 bits or 2 bytes long (reasonable assumption in Null and Lobur especially if you look the previous exercise (No. 4).

4M * 16 = 2^2 * 2^20 * (2^4 / 2^4) (16 bits / 16 bits is a word) = 2^2 * 2^20 * 2^0 = 2^22 => 22 bits.

Upvotes: 5

CompArchFTW
CompArchFTW

Reputation: 81

Lashane is incorrect despite being erroneously upvoted. We're talking about being byte addressable, not bit addressable. In reality, we should end up with 23 bits on a byte addressable system and 22 bits on a word addressable system (assuming the word size is 16 bits wide):

Byte addressable: 4M x 16 = 2^2(4) x 2^20(1m) x 2^1 (2 bytes, or 16 bits) = 2^23, or 23 bits

Word addressable: 4M x 16 = 2^2(4) x 2^20(1m) x 2^0 (1 word, or 16 bits) = 2^22, or 22 bits

Hopefully this clears up any confusion that Lashane may have caused...

Upvotes: 8

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