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# Difference between word logical addressing and byte logical addressing in this exercise?

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Problem Detail:

I´m new here and this is the first question I´m posting (hopefully this is the correct site).

I have the exercise below:
Consider a virtual address space of 10 pages of 1024 words each (1 word = 2 bytes), mapped onto a physical memory of 32 frames. How many bits are needeed for each virtual address?

The answer they provide is:

| 4 bits for page | 10 bits for offset | 

But I don´t figure out how do the 4 bits appear. The exercise explains that the logical addressing is word logical addressing (instead of byte logical addressing).

I´m confused about this two terms. How can I know if the addressing method is word based or byte based?

Thank you in advance

###### Answered By : Maharaj

Page size => 1024 words = 2^10 words

Since it is word addressable format 10 bits will be required to identify offset within page.

There are 10 pages as 10 can't be expressed in 2's power format we need to consider 16 for identifying 10 pages i.e. 2^4=16.

So 4 bits are required for page number. Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/59763

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