How does 32-bit address 4GB if 2³² bits = 4 Billion bits not Bytes?

Question

Essentially, how does 4Gb turn into 4GB? If the memory is addressing Bytes, should not the possibilities be 2^(32/8)?

Answer 1

The electrical interface on the chip consists (extremely simplified) of a wires for the address (e.g. 32 address lines) and wires for the data (e.g. 8 wires for read data coming from the RAM, 8 wires for write data going to the RAM). In this case you have 2^32 words of 8 bits, so you can address 2^32*8 bits of data.

If you had a RAM with a word width of 16bit instead (much more likely than 8bit) you would be able to address twice as much RAM with the same number of address bits. On a modern system, you cannot really "read one byte" but instead the CPU fetches a whole cache line from the RAM and then gives you back just the byte that you asked for.

Answer 2

Most modern computers are byte-addressable, with each address identifying a single eight bit byte of storage; data too large to be stored in a single byte may reside in multiple bytes occupying a sequence of consecutive addresses. There exist word-addressable computers, where minimal addressable storage unit is exactly the processor's word. For example, the Data General Nova minicomputer, and the Texas Instruments TMS9900 and National Semiconductor IMP-16 microcomputers used 16 bit words, and there were many 36-bit mainframe computers (e.g., PDP-10) which used 18-bit word addressing, not byte addressing, giving an address space of 2^18 36-bit words, approximately 1 megabyte of storage. The efficiency of addressing of memory depends on the bit size of the bus used for addresses – the more bits used, the more addresses are available to the computer. For example, an 8-bit-byte-addressable machine with a 20-bit address bus (e.g. Intel 8086) can address 2^20 (1,048,576) memory locations, or one MiB of memory, while a 32-bit bus (e.g. Intel 80386) addresses 2^32 (4,294,967,296) locations, or a 4 GiB address space.

Answer 3

It depends on how you address the data.

If you use 32 bits to address each bit, you can address 2³² bits or 4Gb = 512MB. If you address bytes like most current architectures it will give you 4GB.

But if you address much larger blocks you will need less bits to address 4GB. For example if you address each 512-byte block (2^9 bytes) you can address 4GB with 23 bits. FAT16 uses 16 bits to address (maximum) 64KB clusters and therefore can address a maximum 4GB volume. The same is used in Java Compressed Oops where you can address 32GB of memory with 32-bit reference.

Some older architectures even use word-addressable memory instead of byte like most do nowadays. Modern architectures that have a minimum addressable unit bigger than an octet are mainly found in DSPs. There also a few architectures with bit-addressable memory like Intel 8051