endian http://vvcap.net/db/iTMVDXytoaUXJhefNDJy.png
These two memory spaces store the same data, and I'm trying to write in the counterpart of each one, but I'm confused because, for example, the first given little endian row starts with 11, the lowest number, while the next one starts with 88, the highest. Could anyone please help explain this to me?
As a hint, that's like looking at (decimal) 32 and 57, in one case 3 > 2, but in the other 5 < 7. They're both okay.
The machine is treating each 4-byte row as a single number. The point of endianness is how are those 4 bytes arranged, in terms of memory address. I don't want to be much more explicit and risk depriving you of the exercise, but I find the diagram on wikipedia to be very helpful visualizing this.
http://en.wikipedia.org/wiki/Endianness
Related
I am trying to build a Sieve of Eratosthenes in Lua and i tried several things but i see myself confronted with the following problem:
The tables of Lua are to small for this scenario. If I just want to create a table with all numbers (see example below), the table is too "small" even with only 1/8 (...) of the number (the number is pretty big I admit)...
max = 600851475143
numbers = {}
for i=1, max do
table.insert(numbers, i)
end
If I execute this script on my Windows machine there is an error message saying: C:\Program Files (x86)\Lua\5.1\lua.exe: not enough memory. With Lua 5.3 running on my Linux machine I tried that too, error was just killed. So it is pretty obvious that lua can´t handle the amount of entries.
I don´t really know whether it is just impossible to store that amount of entries in a lua table or there is a simple solution for this (tried it by using a long string aswell...)? And what exactly is the largest amount of entries in a Lua table?
Update: And would it be possible to manually allocate somehow more memory for the table?
Update 2 (Solution for second question): The second question is an easy one, I just tested it by running every number until the program breaks: 33.554.432 (2^25) entries fit in one one-dimensional table on my 12 GB RAM system. Why 2^25? Because 64 Bit per number * 2^25 = 2147483648 Bits which are exactly 2 GB. This seems to be the standard memory allocation size for the Lua for Windows 32 Bit compiler.
P.S. You may have noticed that this number is from the Euler Project Problem 3. Yes I am trying to accomplish that. Please don´t give specific hints (..). Thank you :)
The Sieve of Eratosthenes only requires one bit per number, representing whether the number has been marked non-prime or not.
One way to reduce memory usage would be to use bitwise math to represent multiple bits in each table entry. Current Lua implementations have intrinsic support for bitwise-or, -and etc. Depending on the underlying implementation, you should be able to represent 32 or 64 bits (number flags) per table entry.
Another option would be to use one or more very long strings instead of a table. You only need a linear array, which is really what a string is. Just have a long string with "t" or "f", or "0" or "1", at every position.
Caveat: String manipulation in Lua always involves duplication, which rapidly turns into n² or worse complexity in terms of performance. You wouldn't want one continuous string for the whole massive sequence, but you could probably break it up into blocks of a thousand, or of some power of 2. That would reduce your memory usage to 1 byte per number while minimizing the overhead.
Edit: After noticing a point made elsewhere, I realized your maximum number is so large that, even with a bit per number, your memory requirements would optimally be about 73 gigabytes, which is extremely impractical. I would recommend following the advice Piglet gave in their answer, to look at Jon Sorenson's version of the sieve, which works on segments of the space instead of the whole thing.
I'll leave my suggestion, as it still might be useful for Sorenson's sieve, but yeah, you have a bigger problem than you realize.
Lua uses double precision floats to represent numbers. That's 64bits per number.
600851475143 numbers result in almost 4.5 Terabytes of memory.
So it's not Lua's or its tables' fault. The error message even says
not enough memory
You just don't have enough RAM to allocate that much.
If you would have read the linked Wikipedia article carefully you would have found the following section:
As Sorenson notes, the problem with the sieve of Eratosthenes is not
the number of operations it performs but rather its memory
requirements.[8] For large n, the range of primes may not fit in
memory; worse, even for moderate n, its cache use is highly
suboptimal. The algorithm walks through the entire array A, exhibiting
almost no locality of reference.
A solution to these problems is offered by segmented sieves, where
only portions of the range are sieved at a time.[9] These have been
known since the 1970s, and work as follows
...
I'm taking a microprocessor class and I'm a bit confused on the value of a halfword and word. In lectures, my professor keeps referring to a halfword or a word as though it's a fixed value, like Pi.
My understanding (if correct) is the value depends on the processor, 32bits, 64 bits, etc.. So, for a 32bit processor, a halfword would be 18bits. Would someone clarify this?
As the commenter noted, "word" is a loaded term. Now a days, processors have standardized on processors that use powers of as their "word" size (16, 32, 64, 128). This is why we talk in terms of bytes.
1) This was not so in ye olde days. The PDP-8 was a 12-bit processor. Sperry and the DEC TOPS systems were 36-bits. In these cases a "word" the amount of data the processor inherently worked with: in these examples, 12-bits or 36-bits.
2) In on the PDP-11 and VAX, the term "word" meant 16-bits (even though the VAX was a 32-bit processor).
[There may be other "word" meanings I have omitted.]
Thus a word could mean either the amount of data the processor works with or a specific data length. A half word is half of that amount.
You'd really have to ask to find out what your professor means when he says "word."
Looking at this question on Quora HERE ("Are data stored in registers and memory in hex or binary?"), I think the top answer is saying that data persistence is achieved through physical properties of hardware and is not directly relatable to either binary or hex.
I've always thought of computers as 'binary', but have just realized that that only applies to the usage of components (magnetic up/down or an on/off transistor) and not necessarily the organisation of, for example, memory contents.
i.e. you could, theoretically, create an abstraction in memory that used 'binary components' but that wasn't binary, like this:
100000110001010001100
100001001001010010010
111101111101010100001
100101000001010010010
100100111001010101100
And then recognize that as the (badly-drawn) image of 'hello', rather than the ASCII encoding of 'hello'.
An answer on SO (What's the difference between a word and byte?) mentions that processors can handle 'words', i.e. several bytes at a time, so while information representation has to be binary I don't see why information processing has to be.
Can computers do arithmetic on hex directly? In this case, would the internal representation of information in memory/registers be in binary or hex?
Perhaps "digital computer" would be a good starting term and then from there "binary digit" ("bit"). Electronically, the terms for the values are sometimes "high" and "low". You are right, everything after that depends on the operation. Most of the time, groups of bits are operated on together. Commonly groups are 1, 8, 16, 32 and 64 bits. The meaning of the bits depends on the program but some operations go hand-in-hand with some level of meaning.
When the meaning of a group of bits is not known or important, humans like to be able to decern the value of each bit. Binary could be used but more than 8 bits is hard to read. Although it is rare to operate on groups of 4 bits, hexadecimal is much more readable and is generally used regardless of the number of bits. Sometimes octal is used but that's based on contexts where there is some meaning to a subgrouping of the 3 bits or an avoidance of digits beyond 9.
Integers can be stored in two's complement format and often CPUs have instructions for such integers. Once such operation is negation. For a group of 8 bits, it would map 1 to -1,… 127 to -127, and -1 to 1, … -127 to 127, and 0 to 0 and -128 to -128. Decimal is likely the most valuable to humans here, not base 256, base 2 or base 16. In unsigned hexadecimal, that would be 01 to FF, …, 00 to 00, 80 to 80.
For an intro to how a CPU might do integer addition on a group of bits, see adder circuits.
Other number formats include IEEE-754 floating point and binary-coded decimal.
I think you understand that digital circuits are binary. So, based on the above, yes, operations do operate on a higher conceptual level despite the actual storage.
Not sure if this is a blatantly terrible misunderstanding, but I've been having so trouble with inspecting memory. Here's the following from gdb from examining with x/8w.
0xbffff7a0: 0xb7f9f729 0xb7fd6ff4 0xbffff7d8 0x08048529
0xbffff7b0: 0xb7fd6ff4 0xbffff870 0xbffff7d8 0x00000000
So I'm assuming that 0xb7f9f729 is at 0xbffff7a0, then 0xb7fd6ff4 is at 0xbffff7a4, etc. Could you explain how this works byte wise? Is that 16 bytes from the first memory portion to the next, and each 4 bytes holds it's own word?
I'm having a hard time grasping this memory concept, anyone know a good resource that makes learning it easier?
Yes and yes to both questions.
gdb(1) understands w modifier in your x/8w command as "four byte words", so you are printing 32 bytes in groups of four. gdb(1) just lays them out in short lines with offsets for readability.
I should mention that the exact values printed actually depend on the platform endianness.
You would get similar but probably more understandable layout with x/32.
It's all in the fine manual.
I am preparing for a quiz in my computer science class, but I am not sure how to find the correct answers. The questions come in 4 varieties, such as--
Assume the following system:
Auxiliary memory containing 4 gigabytes,
Memory block equivalent to 4 kilobytes,
Word size equivalent to 4 bytes.
How many words are in a block,
expressed as 2^_? (write the
exponent)
What is the number of bits needed to
represent the address of a word in
the auxiliary memory of this system?
What is the number of bits needed to
represent the address of a byte in a
block of this system?
If a file contains 32 megabytes, how
many blocks are contained in the
file, expressed as 2^_?
Any ideas how to find the solutions? The teacher hasn't given us any examples with solutions so I haven't been able to figure out how to do this by working backwards or anything and I haven't found any good resources online.
Any thoughts?
Questions like these basically boil down to working with exponents and knowing how the different pieces fit together. For example, from your sample questions, we would do:
How many words are in a block, expressed as 2^_? (write the exponent)
From your description we know that a word is 4 bytes (2^2 bytes) and that a block is 4 kilobytes (2^12 bytes). To find the number of words in one block we simply divide the size of a block by the size of a word (2^12 / 2^2) which tells us that there are 2^10 words per block.
What is the number of bits needed to represent the address of a word in the auxiliary memory of this system?
This type of question is essentially an extension of the previous one. First you need to find the number of words contained in the memory. And from that you can get the number of bits required to represent a word in the memory. So we are told that memory contains 4 gigabytes (2^32 bytes) and that the word is 4 bytes (2^2 bytes); therefore the number words in memory is 2^32/2^2 = 2^30 words. From this we can deduce that 30 bits are required to represent a word in memory because each bit can represent two locations and we need 2^30 locations.
Since this is tagged as homework I will leave the remaining questions as exercises :)
Work backwards. This is actually pretty simple mathematics. (Ignore the word "auxilliary".)
How much is a kilobyte? How much is 4 kilobytes? Try putting in some numbers in 2^x, say x == 4. How much is 2^4 words? 2^8?
If you have 4GB of memory, what is the highest address? How large numbers can you express with 8 bits? 16 bits? Hint: 4GB is an even power of 2. Which?
This is really the same question as 2, but with different input parameters.
How many kilobytes is a megabyte? Express 32 megabytes in kilobytes. Division will be useful.