CODEWITHSUNDEEP
bitcpu-architecture
user1534664
user1534664

Reputation: 3418

8 and 16 bit architecture

I'm a bit confused about bit architectures. I just cant find a good article that answers my questions, so I figured I'd ask SO.

Question 1:

When speaking of a 16 bit architecture, does it mean each ram address is 16 bits long? So if I create an int (32 bit) in C++ the variable would take up 2 addresses?

Question 2:

in a 16 bit architecture there are only 2^16 (65536) amount of addresses inside the RAM. Why can't they add more? Is this because 16 bit can't represent a higher value and therefore can't reference to adresses above 65535?

Upvotes: 2

Views: 3298

Answers (4)

KawaiKx
KawaiKx

Reputation: 9940

Example From Wikipedia - All internal registers, as well as internal and external data buses, were 16 bits wide, firmly establishing the "16-bit microprocessor" identity of the 8086. A 20-bit external address bus gave a 1 MB physical address space (2^20 = 1,048,576). This address space was addressed by means of internal 'segmentation'. The data bus was multiplexed with the address bus in order to fit a standard 40-pin dual in-line package. 16-bit I/O addresses meant 64 KB of separate I/O space (2^16 = 65,536). The maximum linear address space was limited to 64 KB, simply because internal registers were only 16 bits wide. Programming over 64 KB boundaries involved adjusting segment registers (see below) and remained so until the 80386 introduced wider (32 bits) registers (and more advanced memory management hardware).

So you can see that there are no fixed rules that a 16 bit architecture will have 16 address lines only. Don't mix up two things, though it's intuitive to believe so.

Upvotes: 1

JasonD
JasonD

Reputation: 16612

Depends on your definitions of 8-bit and 16-bit architecture.

The 6502 was considered an 8-bit CPU, because it operated on 8-bit values (the register size), yet had 16-bit addresses.

The 68000 was considered a 16-bit CPU, yet had 32-bit registers and addresses.

With x86, it is generally the address size that defines the architecture.

Also, '64-bit' CPUs don't always have a full 64-bit external address bus. They might internally handle addresses of that size, so the virtual address space can be large, but it doesn't mean they can have that much external memory.

Upvotes: 1

David Schwartz
David Schwartz

Reputation: 182837

When speaking of a 16 bit architecture, does it mean each ram address is 16 bits long? So if I create an int (32 bit) in C++ the variable would take up 2 addresses?

You'd have to ask whoever was speaking of a 16-bit architecture what they meant by it. They could mean addresses are 16-bits long. They could mean general-purpose CPU registers are 16-bits long. They could mean something else. But there's no way we could know what some hypothetical person might mean. There is no universal definition of what makes something a "16-bit architecture".

For example, the 8032 is an 8-bit architecture with 8-bit general purpose registers. But it has a 16-bit pointer register that can be used to address 65,536 bytes of storage.

Regardless of bitness, almost all systems use byte addresses. So a 32-bit variable will take up 4 addresses on a machine of any bitness.

in a 16 bit architecture there are only 2^16 (65536) amount of addresses inside the RAM. Why can't they add more? Is this because 16 bit can't represent a higher value and therefore can't reference to adresses above 65535?

With 16-bits, there are only 65,536 possible ways those bits can be set. So a 16-bit register has 65,536 possible values.

Upvotes: 2

SLaks
SLaks

Reputation: 888047

  1. Yes. Note, though that int on 16-bit architectures is usually just 16 bits wide.
    Also note that it doesn't make sense to say that a variable "takes up" two addresses. The correct thing to say is that a 32-bit variable is as wide as two pointers on a 16-bit platform.
    It will still occupy four bytes of space, no matter what architecture.

  2. Yes; that's exactly what 16-bit addresses mean.
    Note that each of these addresses points to a single byte of memory.

Upvotes: 1

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