Consider a machine with a byte addressable main memory of 2^32 bytes divided…
2017
Consider a machine with a byte addressable main memory of 2^32 bytes divided into blocks of size 32 bytes. Assume that a direct mapped cache having 512 cache lines is used with this machine. The size of the tag field in bits is _______
Attempted by 303 students.
Show answer & explanation
Correct answer: 18
Answer: 18 bits.
Explanation:
Total physical address size is 32 bits (2^32 bytes of memory).
Block (offset) bits = log2(block size) = log2(32) = 5 bits.
Index bits = log2(number of cache lines) = log2(512) = 9 bits.
Tag bits = address bits − (index bits + offset bits) = 32 − (9 + 5) = 18 bits.
A video solution is available for this question — log in and enroll to watch it.