Question

Consider a processor that generates 32-bit virtual addresses. It uses a direct mapped cache whose size is 32 KByte and has 32 Byte blocks. What is the number of bits needed for cache addressing?

• A. 5
• B. 10
• C. 15
• D. 20

Hint:

Cache addressing bits are calculated as: Index bits + Block offset bits.

Answer 1

The Correct Option is C

Step 1: The number of bits needed to address every byte inside the cache is simply \( \log_2(\text{cache size in bytes}) \), since each unique bit pattern must map to one byte location.
Step 2: Cache size = 32 KB = \(2^{15}\) bytes, so the number of address bits required is \( \log_2(2^{15}) = 15 \).
Step 3: This total naturally splits into 10 bits for choosing the block (since there are \(2^{10}\) blocks of 32 bytes each) and 5 bits for the offset within that block, confirming the same total.
\[ \boxed{15} \]

Answer 2

Another way to settle this is by checking how many distinct locations each option's bit-count can actually address, using \(2^{\text{bits}}\), and comparing that to the cache's real byte count of 32 KB = 32768 bytes:

1. 5 bits: \(2^5 = 32\) addressable locations, enough only for one 32-byte block, far short of the full cache.
2. 10 bits: \(2^{10} = 1024\) addressable locations, this matches the number of blocks (1024), not the number of bytes, so it under-counts the addressing needed for byte-level access.
3. 15 bits: \(2^{15} = 32768\) addressable locations, which exactly equals the cache size of 32 KB in bytes. Every byte in the cache gets a unique address with exactly this many bits.
4. 20 bits: \(2^{20} = 1{,}048{,}576\) addressable locations, this is far larger than the 32 KB cache actually contains, so it massively overshoots what's needed.

Matching \(2^{\text{bits}}\) to the actual cache capacity singles out one option cleanly.

Therefore, the correct answer is 15.