Question

A byte addressable direct-mapped cache has 1024 blocks/lines, with each block having eight 32-bit words. How many bits are required for block offset, assuming a 32-bit address

Answer 1

Answer and Explanation:

"The inquiry as presented is not necessarily responsible. A word has been states as 32-bit. We need to ask if the frame is "byte-addressable" (From this we can access to get an 8-bit piece of information) or "text-addressable" (the smallest open lump is 32-bit) or maybe "half-word-addressable" (the tiny bundle of information it could reach to 16-bit).

To understand what the smallest request bit of a position is to let anyone know, you have to remember this.

You operate from base up at that stage. We will agree with the byte-addressable structure. Every reserved square at a certain point contains 8 words * (4 bytes/word) = 32 = 25 bytes, so the counterbalance seems to be 5 bits.

The history in a direct-mapped stored is the squares in reserves (12 bits for this position due to 212 = 4096). at a certain point, as you have seen, the tag is also one of the bits left behind.

As the reserve becomes increasingly cooperative. And a similar size remains. These are lesser bits on the list and more bits on the mark.'