If a computer system has a cache memory of 256KB and a main memory of 4GB, what is the number of bits required to address the main memory?
Q1. If a computer system has a cache memory of 256KB and a main memory of 4GB, what is the number of bits required to address the main memory?
Answer: 32 bits
Explanation: Number of bits required to address main memory = log2(main memory size) = log2(4GB) = log2(2³2) = 32