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the upper leftmost 7 bits, so the tags are different and unique.To see how these addresses differ, let’s look at a smaller, simpler example.Suppose we have a system using direct mapping with 16 words of main memorydivided into 8 blocks (so each block has 2 words). Assume the cache is 4 blocksin size (for a total of 8 words). Table 6.1 shows how the main memory blocksmap to cache.We know:• A main memory address has 4 bits (because there are 24 or 16 words inmain memory).• This 4-bit main memory address is divided into three fields: The word field is 1 bit(we need only 1 bit to differentiate between the two words in a block); the blockfield is 2 bits (we have 4 blocks in main memory and need 2 bits to uniquely identifyeach block); and the tag field has 1 bit (this is all that is left over).The main memory address is divided into the fields shown in Figure 6.6.
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