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Data bus width and word size
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How do we determine if a processor is 8-bit; 16-bit or 32-bit
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word size and data bus
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Well, I just recently started reading the book: Structured Computer Organization, By Andrew Tannerbaun, and everthing was clear to me until I reached this sentence on ch.2: "Finally, many computers can transfer 64 or 128 bits in parallel on a single bus cycle, even on 32-bit machines". The problem with this is that I cannot picture how something like this would work and, as far as I know, a cpu has a single data bus.
If there were for example, a 32bit CPU in a 64bit system (64bit data bus), how would the CPU do to transfer the 64bits "in parallel" on the same bus cycle?
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How to get the number of actual cores on the cpu on windows? [duplicate]
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How to Detect the Number of Physical Processors / Cores on Windows, Mac and Linux
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I tried this but it will shows number of logical processors only
SYSTEM_INFO sysinfo;
GetSystemInfo(&sysinfo);
int numCPU = sysinfo.dwNumberOfProcessors;
From https://msdn.microsoft.com/en-us/library/windows/desktop/ms724958(v=vs.85).aspx:
Note For information about the physical processors shared by logical processors, call GetLogicalProcessorInformationEx with the RelationshipType parameter set to RelationProcessorPackage (3).
You can get the related hardware of the logical processors, and infer how many physical processors are there
I got a doubt about the width of internal data bus of AVR controllers connected to flash memory. I was mainly referring to Atmega328. Datasheet says (Page 17) "Since all AVR instructions are 16 or 32 bits wide, the Flash is organized as
2/4/8/16K x 16.". That means flash memory data bus width must be 16 bit? I could not see anywhere mentioning about 16 bit wide program memory data bus (Of course internal to the controller). But bus for RAM seems to be again 8 bit. Just want a clarification.
BitThe 8-bit AVR family is based on a (modified) Harvard architecture, where you have dedicated program and data storages. The data path to program memory is indeed 16-bit, while it is 8-bit only to data memory.
The funny part is, that in the beginning Atmel points out, that these are 8-bit CPUs. This makes them look very competitive when compared to other 8-bit products like 8051 or Rabbit. Due to the 16-bit program data path the AVRs perform very well in benchmark tests. Later, when 8-bit sounds a bit old-fashioned, Atmel decided to call them 8/16-bit CPUs.
Figure 7.1 on page 9 of the data sheet/complete shows that the flash isn't at all connected to the (8 bit) data bus but only to an address bus. The "data" of the flash memory primarily goes into the instruction register and by use of the LPM instruction this data is transferred into a register. Note that when writing data to the flash you always write 16 bit (R1:R0) addressed by the Z pointer (SPM instruction) ... and that the SPM instruction cannot be expressed in "clock cycles" (pg. 617)
I was wondering how exactly does a CPU request data in a computer. In a 32 Bits architecture, I thought that a computer would put a destination on the address bus and would receive 4 Bytes on the data bus. I recently read on the memory alignment in computer and it confused me. I read that the CPU has to read two times the memory to access a not multiple 4 address. Why is so? The address bus lets it access not multiple 4 address.
The address bus itself, even in a 32-bit architecture, is usually not 32 bits in size. E.g. the Pentium's address bus was 29 bits. Given that it has a full 32-bit range, in the Pentium's case that means each slot in memory is eight bytes wide. So reading a value that straddles two of those slots means two reads rather than one, and alignment prevents that from happening.
Other processors (including other implementations of the 32-bit Intel architecture) have different memory access word sizes but the point is generic.
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Possibly the most basic computer question of all time, but I have not been able to find a straightforward answer and it is driving me crazy. When a computer 'reads' a byte, does it read it as a sequential series of ones and zeros one after the other, or does it somehow read all 8 ones and zeros at once?
A computer system reads the data in both ways depending upon type of operation and the how the digital system is designed.I'll explain this with very simple example of a Full adder circuit.
A full adder adds binary numbers and accounts for values carried in as well as out (Wikipedia)
Example of Parallel operation
Suppose in some task we need to add two 8 bit(1 byte) numbers such that all bits are available at the time of addition.
Then in that case we can design a digital system with 8 full-adders(1 for each bit).
Example of Serial Operation
In some other task you observe that all 8 bits will not be simultaneously available.
Or you think having 8 separate adders is costly as you need to implement other mathematical operations (like subtraction,multiplication and division). So instead of having 8 separate units you have 1 unit which will individually process bits. In this scenario we will need three storage units ( Shift Registers) such that two storage units will store two 8-bit numbers and one storage units will store the result .At a given clock pulse single bit will be transmitted from each of two registers to the full adder which will perform the addition process and transfer 1 bit result to the result shift register in single clock pulse.
This figure contains some additional stuff which is not useful for this thread but you can
study digital logic design and computer architecture if you want to go more deep in this stuff.
Shift register
Shift register operations demo
This is really kind of outside the scope of Stackoverflow, but it brings back such fond memories from college.
It depends. Some times a computer reads bits one at a time. For example over older ethernet manchester code is used. However over old parallel printer cables, there were 8 pins each one signaling a bit, and an entireoctet (byte) is sent at once.
In serial (one-bit-at-a-time) encodings, you're typically measuring transitions in the line or transitions against some well-defined clock source.
In parallel encodings, you're typically reading all the bits into a register at a time and latching the register.
Look up flipflops, registers, and logic gates for information on the low-level parts of this.
Bits are transmitted one at a time in serial transmission, and
multiple numbers of bits in parallel transmission. A bitwise operation
optionally process bits one at a time. Data transfer rates are usually
measured in decimal SI multiples of the unit bit per second (bit/s),
such as kbit/s.
Wikipedia's article on Bit
the processor works with a defined number of registerlength. 8, 16, 32, 64 ... think about a register as an amount of connection, one for each bit... thats the amount of bits that will be processed at once in one processor core, one register at once ... the processor hat different kinds of register, examples are the private instruction register or the public data or adress register
Think of it this way, at least at a physical level: In a transmission cable from point A to B (A and B can be anything, hard drive, CPU, RAM, USB, etc.) each wire in that cable can transmit one bit at a time. Both A and B have a clock pulsing at the same rate. On each pulse, the sender changes the amount of power going down each wire to signify the value of the new bit(s). So, the # of wires in the cable = the # of bits that can be transmitted each "pulse". (Note: This is a very simplified and theoretical explanation).
At a software level, in the CPU, you can never address anything smaller than a byte. You can "access" and manipulate specific bytes by using the bitwise operators (& (AND), | (OR), << (Left Shift), >> (Right Shift), ^ (XOR)).
In hardware, the number of bits being sent each pulse is completely dependent of the hardware itself.
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.