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Difference between word addressable and byte addressable

Can someone explain what's the different between Word and Byte addressable? How is it related to memory size etc.?

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leon Avatar asked Apr 27 '10 19:04

leon


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What is difference between byte and word?

A byte is eight bits, a word is 2 bytes (16 bits), a doubleword is 4 bytes (32 bits), and a quadword is 8 bytes (64 bits). Figure 29-2 shows the byte order of each of the fundamental data types when referenced as operands in memory.

What do you mean by byte-addressable?

The 386SX, which addresses memory in 8-bit units but can fetch and store it 16 bits at a time, is termed byte-addressable. The advantage of word addressing is that more memory can be addressed in the same number of bits.

What is difference between bit addressable and byte-addressable instruction?

They are not really using the terms right, byte addressable is what we are used to an address represents a unique byte in memory or the memory space. Bit addressable would mean that each bit in the memory space has a unique address, which is not the case.

Is x86 byte-addressable or word-addressable?

The x86 Intel architecture is also byte addressable. If we have an integer-sized piece of data, we may want to identify the various bits. This can be done by numbering the bits.


1 Answers

  • A byte is a memory unit for storage
  • A memory chip is full of such bytes.

Memory units are addressable. That is the only way we can use memory.

In reality, memory is only byte addressable. It means:

  • A binary address always points to a single byte only.
  • A word is just a group of bytes2, 4, 8 depending upon the data bus size of the CPU.

To understand the memory operation fully, you must be familiar with the various registers of the CPU and the memory ports of the RAM. I assume you know their meaning:


  • MAR(memory address register)
  • MDR(memory data register)
  • PC(program counter register)
  • MBR(memory buffer register)

RAM has two kinds of memory ports:

  1. 32-bits for data/addresses
  2. 8-bit for OPCODE.

Suppose CPU wants to read a word (say 4 bytes) from the address xyz onwards. CPU would put the address on the MAR, sends a memory read signal to the memory controller chip. On receiving the address and read signal, memory controller would connect the data bus to 32-bit port and 4 bytes starting from the address xyz would flow out of the port to the MDR.

If the CPU wants to fetch the next instruction, it would put the address onto the PC register and sends a fetch signal to the memory controller. On receiving the address and fetch signal, memory controller would connect the data bus to 8-bit port and a single byte long opcode located at the address received would flow out of the RAM into the CPU's MDR.

So that is what it means when we say a certain register is memory addressable or byte addressable. Now what will happen when you put, say decimal 2 in binary on the MAR with an intention to read the word 2, not (byte no 2)?

Word no 2 means bytes 4, 5, 6, 7 for 32-bit machine. In real physical memory is byte addressable only. So there is a trick to handle word addressing.

When MAR is placed on the address bus, its 32-bits do not map onto the 32 address lines(0-31 respectively). Instead, MAR bit 0 is wired to address bus line 2, MAR bit 1 is wired to address bus line 3 and so on. The upper 2 bits of MAR are discarded since they are only needed for word addresses above 2^32 none of which are legal for our 32 bit machine.
Using this mapping, when MAR is 1, address 4 is put on the bus, when MAR is 2, address 8 is put on the bus and so forth.

It is a bit difficult in the beginning to understand. I learnt it from Andrew Tanenbaums's structured computer organisation.

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KawaiKx Avatar answered Oct 16 '22 12:10

KawaiKx