Cache and Frames Question

Posted on 2004-03-24
Last Modified: 2010-04-26
Can someone explain to me the following:

I found this notes on the web:

suppose we access the memory at address 0x0023AF7C. The looking at that in binary, that is 000000000010001110101111011111002. If we separate these bits into the lengths of the fields we have determined with dashes between each field, that is more easily read as 00000000001000111010111-1011-11100. So, our index is 10112=1110. We look at index 11. Is there anything in there yet? We see there is not. Therefore, we load the data contained at memory addresses 0x0023AF60 through 0x0023AF7F into the 32 byte line of the cache block with index 11.

okay, what i don understand is..., this is obviousl a miss, so we need to get the data from the main memory, but why it says here that we load the data at memory 0x0023AF60 through 0x0023AF7F into....??? Shoudnt we just load it from the address 0x0023AF7C??? Where are those 0x023AF60 and 0x0023AF7F came from??????
Question by:jtcy
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Accepted Solution

Callandor earned 125 total points
ID: 10673994
I think the way memory works is it is moved a block at a time, not a byte at a time.  This is done for performance reasons because if you are reading memory, you usually get a contiguous block at a time.

Author Comment

ID: 10674040
yes, but isnt 0x0023AF7C a block?
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Expert Comment

ID: 10674075
No, it's an address - the range 0x0023AF60 through 0x0023AF7F is a block (and as you noted, it's a cache block).
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Author Comment

ID: 10674097
cool. i understand now
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Expert Comment

ID: 10674116
Glad I could help.

Author Comment

ID: 10674293
Can i just ask one more, cache doesnt store the real data right? The data it stores is actually the memory address of the real data, am i right?
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Expert Comment

ID: 10674358
No, cache would be of no use if it didn't store real data.  The idea is to save reading it from main memory, which is slower than cache memory.

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