For example, for a 8-bit CPU, the stack size is expected to be 8-bit wide, and 16-bit CPU vs 16-bit stack width, and 32-bit, 64-bit CPU, and so on. Is it true for all architectures?
A CPU has a data bus and a address bus. They can have the same width, but they often aren't.
The stack pointer is a pointer to memory, so its often as wide as the address bus, unless there is some (weird/obscure) conversion used internally. The instruction pointer (points to the current instruction) is also a pointer to memory and as such as wide as the stack pointer.
Other registers mostly deal with data, and as such have the same dimensions as the data bus. But as usual, there are exceptions.
To take an old example. The 6502. A 8 bit cpu (8 bit databus, 16 bit addressbus). It has the (more or less) general purpose registers X and Y, and the "accumulator" called A. All 8 bit registers. There is a stack pointer and a instruction pointer, the stack pointer having 8 explicit and 8 implicit bits (stack is always in the same 256-byte region) and thus the stack pointer register having 8 bits, the instruction pointer having 16 bits.
The 8086, has an 16 bit databus and a 20 bit addressbus. The general registers where 8 (and 16 bit). The instruction pointer and stackpointer where 16 bit, but used segment registers (also 16 bit) to get the full 20 bit address.
8-bit CPU, the stack size would expect
to be 8-bit width
You might expect that, but you would be wrong. The basic use of the stack is to store & retrieve return addresses, and on 8-bit processors these are actually 16-bit values. For example the Z80 instruction:
PUSH HL
pushes a 16-bit value (the contents of the two 8-bit registers, H and L) onto the stack. There is no Z80 instruction which pushes an 8-bit value.
The stack is really just a concept, and "stack width" doesn't really mean anything.
Presumably, you mean the width of the stack elements, but you can typically store arbitrary sized values on any stack (with some padding or alignment), regardless of the size of the architecture's bus width or register size.
Note that the terms 8-bit, 16-bit etc are a little arbitrary, and aren't really tight definitions, so this complicates any attempted definition of "width".
One point not yet mentioned is that some CPUs (e.g. the Microchip PIC family) have a dedicated stack which is not accessed like normal RAM. On the PIC16 series, the stack is tied exclusively to the program counter, and the stack is (presumably) the same width as the program counter. On the PIC18 series, the stack and program counter are both 21 bits wide; the upper 20 bits can be copied to or from the program counter; code can also examine the top stack element as a 5-bit part and two 8-bit parts.
Related
My teacher has given me the question to differentiate the maximum memory space of 1MB and 4GB microprocessor. Does anyone know how to answer this question apart from size mentioned difference ?
https://i.stack.imgur.com/Q4Ih7.png
A 32-bit microprocessor can address up to 4 GB of memory, because its registers can contain an address that is 32 bits in size. (A 32-bit number ranges from 0 to 4,294,967,295). Each of those values can represent a unique memory location.
The 16-bit 8086, on the other hand, has 16-bit registers which only range from 0 to 65,535. However, the 8086 has a trick up its sleeve- it can use memory segments to increase this range up to one megabyte (20 bits). There are segment registers whose values are automatically bit-shifted left by 4 then added to the regular registers to form the final address.
For example, let's look at video mode 13h on the 8086. This is the 256-color VGA standard with a resolution of 320x200 pixels. Each pixel is represented by a single byte and the desired color is stored in that byte. The video memory is located at address 0xA0000, but since this value is greater than 16 bits, typically the programmer will load 0xA000 into a segment register like ds or es, then load 0000 into si or di. Once that is done, the program can read from [ds:si] and write to [es:di] to access the video memory. It's important to keep in mind that with this memory addressing scheme, not all combinations of segment and offset represent a unique memory location. Having es = A100/di = 0000 is the same as es=A000/di=1000.
Why is the register length (in bits) that a CPU operates on not dynamically/manually/arbitrarily adjustable? Would it make the computer slower if it was adjustable this way?
Imagine you had an 8-bit integer. If you could adjust the CPU register length to 8 bits, the CPU would only have to go through the first 8 bits instead of extending the 8-bit integer to 64 bits and then going through all 64 bits.
At first I thought you were asking if it was possible to have a CPU with no definitive register size. That make no sense since the number and size of the registers is a physical property of the hardware and cannot be changed.
However some architecture let the programmer work on a smaller part of a register or to pair registers.
The x86 does both for example, with add al, 9 (uses only 8 bits of the 64-bit rax) and div rbx (pairs rdx:rax to form a 128-bit register).
The reason this scheme is not so diffuse is that it comes with a lot of trade-offs.
More registers means more bits needed to address them, simply put: longer instructions.
Longer instructions mean less code density, more complex decoders and less performance.
Furthermore most elementary operations, like the logic ones, addition and subtraction are already implemented as operating on a full register in a single cycle.
Finally, one execution unit can handle only one instruction at a time, we cannot issue eight 8-bit additions in a 64-bit ALU at the same time.
So there wouldn't be any improvement, nor in the latency nor in the throughput.
Accessing partial registers is useful for the programmer to fan-out the number of available registers, so for example if an algorithm works with 16-bit data, the programmer could use a single physical 64-bit register to store four items and operate on them independently (but not in parallel).
The ISAs that have variable length instructions can also benefit from using partial register because that usually means smaller immediate values, for example and instruction that set a register to a specific value usually have an immediate operand that matches the size of register being loaded (though RISC usually sign-extends or zero-extends it).
Architectures like ARM (presumably others as well) supports half precision floats. The idea is to do what you were speculating and #Margaret explained. With half precision floats, you can pack two float values in a single register, thereby introducing less bandwidth at a cost of reduced accuracy.
Reference:
[1] ARM
[2] GCC
On a x86 system a memory location can hold 4 bytes (32 / 8) of data, therefore a single memory address in a 64 bit system can hold 8 bytes per memory address. When examining the stack in GDB though this doesn't appear to be the case, example:
0x7fff5fbffa20: 0x00007fff5fbffa48 0x0000000000000000
0x7fff5fbffa30: 0x00007fff5fbffa48 0x00007fff857917e1
If I have this right then each hexadecimal pair (48) is a byte, thus the first memory address
0x7fff5fbffa20: is actually holding 16 bytes of data and not 8.
This has had me really confused and has for a while, so absolutely any input is vastly appreciated.
Short answer: on both x86 and x64 the minimum addressable entity is a byte: each "memory location" contains one byte, in each case. What you are seeing from GDB is only formatting: it is dumping 16 contiguous bytes, as the address increasing from ....20 to ....30, (on the left) indicates.
Long answer: 32bit or 64bit is used to indicate many things, in an architecture: almost always, is the addressable size (how many bits are in an address = how much memory you can directly address - again, bytes of memory). It also usually indicates the dimension of registers, and also (but not always) the native word size.
That means that usually, even if you can address a single byte, the machine works "better" using data of different (longer) size. What "better" means is beyond the question; a little background, however, is good to understand some misconceptions about word size in the question.
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?
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.
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.
Yes; that's exactly what 16-bit addresses mean.
Note that each of these addresses points to a single byte of memory.
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.
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.
I heard that the 8086 has 16-bit registers which allow it to only address 64K of memory. Yet it is still able to address 1MB of memory which would require 20-bit registers. It does this by using another register to to hold another 16 bits, and then adds the value in the 16-bit registers to the value in this other register to be able to generate numbers which can address up to 1MB of memory. Is that right?
Why is it done this way? It seems that there are 32-bit registers, which is more than sufficient to address 1MB of memory.
Actually this has nothing to do with number of registers. Its the size of the register which matters. A 16 bit register can hold up to 2^16 values so it can address 64K bytes of memory.
To address 1M, you need 20 bits (2^20 = 1M), so you need to use another register for the the additional 4 bits.
The segment registers in an 8086 are also sixteen bits wide. However, the segment number is shifted left by four bits before being added to the base address. This gives you the 20 bits.
the 8088 (and by extension, 8086) is instruction compatible with its ancestor, the 8008, including the way it uses its registers and handles memory addressing. the 8008 was a purely 16 bit architecture, which really couldn't address more than 64K of ram. At the time the 8008 was created, that was adequate for most of its intended uses, but by the time the 8088 was being designed, it was clear that more was needed.
Instead of making a new way for addressing more ram, intel chose to keep the 8088 as similar as possible to the 8008, and that included using 16 bit addressing. To allow newer programs to take advantage of more ram, intel devised a scheme using some additional registers that were not present on the 8008 that would be combined with the normal registers. these "segment" registers would not affect programs that were targeted at the 8008; they just wouldn't use those extra registers, and would only 'see' 16 addres bits, the 64k of ram. Applications targeting the newer 8088 could instead 'see' 20 address bits, which gave them access to 1MB of ram
I heard that the 8086 has 16 registers which allow it to only address 64K of memory. Yet it is still able to address 1MB of memory which would require 20 registers.
You're misunderstanding the number of registers and the registers' width. 8086 has eight 16-bit "general purpose" registers (that can be used for addressing) along with four segment registers. 16-bit addressing means that it can only support 216 B = 64 KB of memory. By getting 4 more bits from the segment registers we'll have 20 bits that can be used to address a total of 24*64KB = 1MB of memory
Why is it done this way? It seems that there are 32 registers, which is more than sufficient to address 1MB of memory.
As said, the 8086 doesn't have 32 registers. Even x86-64 nowadays don't have 32 general purpose registers. And the number of registers isn't relevant to how much memory a machine can address. Only the address bus width determines the amount of addressable memory
At the time of 8086, memory is extremely expensive and 640 KB is an enormous amount that people didn't think that would be reached in the near future. Even with a lot of money one may not be able to get that large amount of RAM. So there's no need to use the full 32-bit address
Besides, it's not easy to produce a 32-bit CPU with the contemporary technology. Even 64-bit CPUs today aren't designed to use all 64-bit address lines
Why can't OS use entire 64-bits for addressing? Why only the 48-bits?
Why do x86-64 systems have only a 48 bit virtual address space?
It'll takes more wires, registers, silicons... and much more human effort to design, debug... a CPU with wider address space. With the limited transistor size of the technology in the 70s-80s that may not even come into reality.
8086 doesn't have any 32-bit integer registers; that came years later in 386 which had a much higher transistor budget.
8086's segmentation design made sense for a 16-bit-only CPU that wanted to be able to use 20-bit linear addresses.
Segment registers could have been only 8-bit or something with a larger shift, but apparently there are some advantages to fine-grained segmentation where a segment start address can be any 16-byte aligned linear address. (A linear address is computed from (seg << 4) + off.)