DIRECT CACHE

Question

https://gateoverflow.in/863/gate2002-10

IN THE SOLUTION BY ARJUN SIR HOW THIS LINE IS JUSTIFYING THAT ALL WILL MAP TO DIFFERENT SET??????

 So, since the ending address is not extending beyond these 9 bits,9 bits, all cache accesses are to diff sets. 

Comments

Array contains 2048 elements each of size 4 bytes, block size of cache is 16 bytes.  so in each cache block we can store up to 4 elements.

Now how many blocks are needed to store 2048 elements =  2048/4 = 512 blocks are needed. It means I can store 2048 elements in 512 cache blocks.

how many cache block we can have = 8K/16 = 512, also memory address is starting from 0000....  to store array elements hence mod 512 will be from 0 to 511

Hence we need 512 blocks to store 2048 and even cache contains 512 blocks, so our last array element will be placed in last cache block. Hence He can say ending address of the array is not extending beyond the 9 bits. if there would have been 2049 elements then we had to map 2049^th element in first cache block. How and why? 0-2048 elements will be placed in blocks 0-511 hence total of 512 blocks. now 2049 elements will give block address of 512 hence 512 mod 512 = 0 Hence we can place 2049th element in first cache block. Hence using only 2048 we can say all cache access are to different sets and not repeated.

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OK GOT IT