Why is the address space of a 32-bit processor 4Gibibytes and not 4Gibibits?

Question

A 32-bit register can store 2³² different values. The signed range of integer values that can be stored in 32 bits is -2,147,483,648 through 2,147,483,647 (unsigned: 0 through 4,294,967,295). Hence, a processor with 32-bit memory addresses can directly access 4 GiB of byte-addressable memory.

https://en.wikipedia.org/wiki/32-bit

What confuses me is that we are talking about a 32-bit processor which can address 2³² different addresses. But now 2³² is in bytes and not bits, why is that?

Because you can't typically address anything smaller than a byte in memory. If you need a particular bit, read a byte/halfword/word/whatever, and extract the bit you need using bitwise logic. — Michael, Aug 23 '15 at 19:54

score 2 · Accepted Answer · answered Aug 23 '15 at 20:06

The smallest addressable unit in memory is a byte (exceptions to this rule exist, but you were asking about x86 specifically). So if your addresses are N-bit, you can represent 2^N different addresses, which means that you can address 2^N individual bytes.

score 1 · Answer 2 · edited Oct 28 '18 at 18:35

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Each unique address points to a byte of memory in the address map, and not a bit. In other words, memory is byte addressable that's why 32bits can address 4GBytes.

edited Oct 28 '18 at 18:35

Sunil Singh

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answered Oct 28 '18 at 17:42

Anand K Enamandram

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Why is the address space of a 32-bit processor 4Gibibytes and not 4Gibibits?

2 Answers2